IGNOU BMTC-133 Solved Assignment | 1st January, 2025 to 31st December, 2025 | REAL ANALYSIS | BSc CBCS

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IGNOU BMTC-133 Solved Assignment | 1st January, 2025 to 31st December, 2025 | REAL ANALYSIS | BSc CBCS Sample Solution

BMTC-133 Solved Assignment 2025

  1. Which of the following statements are true or false? Give reasons for your answers in the form of a short proof or counter-example, whichever is appropriate:
    i) Every infinite set is an open set.
    ii) The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q.
    iii) -1 is a limit point of the interval ] − 2 , 1 ] ] − 2 , 1 ] ]-2,1]]-2,1]]−2,1].
    iv) The necessary condition for a function f f fff to be integrable is that it is continuous.
    v) The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x)=|x-2|+|3-x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x=5x=5.
  2. a) Test the following series for convergence.
    (i) ∑ n − 1 ∞ n x n − 1 , x > 0 ∑ n − 1 ∞   n x n − 1 , x > 0 quadsum_(n-1)^(oo)nx^(n-1),x > 0\quad \sum_{\mathrm{n}-1}^{\infty} \mathrm{n} \mathrm{x}^{\mathrm{n}-1}, \mathrm{x}>0∑n−1∞nxn−1,x>0.
    (ii) ∑ n = 1 ∞ [ n 4 + 9 − n 4 − 9 ] ∑ n = 1 ∞   n 4 + 9 − n 4 − 9 sum_(n=1)^(oo)[sqrt(n^(4)+9)-sqrt(n^(4)-9)]\sum_{n=1}^{\infty}\left[\sqrt{n^4+9}-\sqrt{n^4-9}\right]∑n=1∞[n4+9−n4−9]
    b) Prove that the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_n\right)_{n \in \mathbb{N}}(an)n∈N, where a n = 3 2 x 2 + 2 2 a n = 3 2 x 2 + 2 2 a_(n)=(3^(2))/(x^(2)+2^(2))a_n=\frac{3^2}{x^2+2^2}an=32x2+22, is Cauchy.
  3. a) Show that the set:
S = { 1 n + 1 + ( − 1 ) n 3 : n ∈ N } S = 1 n + 1 + ( − 1 ) n 3 : n ∈ N S={(1)/(n)+(1+(-1)^(n))/(3):n inN}S=\left\{\frac{1}{n}+\frac{1+(-1)^n}{3}: n \in \mathbb{N}\right\}S={1n+1+(−1)n3:n∈N}
is not closed.
b) Test the following series for convergence:
1 3.4 + 2 5.6 + 3 7.8 + … … . . to ∞ 1 3.4 + 2 5.6 + 3 7.8 + … … . . to ∞ (1)/(3.4)+(sqrt2)/(5.6)+(sqrt3)/(7.8)+dots dots..” to “oo\frac{1}{3.4}+\frac{\sqrt{2}}{5.6}+\frac{\sqrt{3}}{7.8}+\ldots \ldots . . \text { to } \infty13.4+25.6+37.8+…….. to ∞
  1. a) Prove that a function f : S → S f : S → S f:S rarr Sf: S \rightarrow Sf:S→S (where S S SSS is a finite non-empty set) is injective if it is surjective.
    b) Disprove the statement:
" ( x + y ) n = x n + y n ∀ n ∈ N , x , y ∈ Z " " ( x + y ) n = x n + y n ∀ n ∈ N , x , y ∈ Z " “(x+y)^(n)=x^(n)+y^(n)AA n inN,x,y inZ^(“)“(x+y)^n=x^n+y^n \forall n \in \mathbb{N}, x, y \in \mathbb{Z}^”“(x+y)n=xn+yn∀n∈N,x,y∈Z“
by providing a suitable counter-example.
c) Find the supremum and infimum of the set:
S = { 1 n − 1 : n ≥ 2 } . S = 1 n − 1 : n ≥ 2 . S={(1)/(n-1):n >= 2}.S=\left\{\frac{1}{n-1}: n \geq 2\right\} .S={1n−1:n≥2}.
  1. a) Show that [ a , ∞ ) [ a , ∞ ) [a,oo)[a, \infty)[a,∞) is closed set.
    b) Write the following statement, and its negation, using logical quantifiers. Also interpret its negation in words.
    ∃ x ∈ R ∃ x ∈ R EE x inR\exists x \in \mathbb{R}∃x∈R such that x − 1 3 > 0 x − 1 3 > 0 x-(1)/(3) > 0x-\frac{1}{3}>0x−13>0.
    c) Let x x xxx and y y yyy be two real numbers such that x < y x < y x < yx<yx<y. Show that there exists an irrational number λ λ lambda\lambdaλ such that
x < λ < y . x < λ < y . x < lambda < y.x<\lambda<y .x<λ<y.
  1. a) Prove that between any two real roots of e x cos ⁡ 2 x = 2 e x cos ⁡ 2 x = 2 e^(x)cos 2x=2e^x \cos 2 x=2excos⁡2x=2, there is at least one real root of
e x sin ⁡ 2 x = 1 . e x sin ⁡ 2 x = 1 . e^(x)sin 2x=1″. “e^x \sin 2 x=1 \text {. }exsin⁡2x=1.
b) Let f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R be a function defined by:
f ( x ) = { x 3 cos ⁡ ( 1 2 x ) , if x ≠ 0 0 , if x = 0 f ( x ) = x 3 cos ⁡ 1 2 x , if x ≠ 0 0 , if x = 0 f(x)={[x^(3)cos((1)/(2x))”,”,” if “x!=0],[0″,”,” if “x=0]:}f(x)=\left\{\begin{array}{cc} x^3 \cos \left(\frac{1}{2 x}\right), & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.f(x)={x3cos⁡(12x), if x≠00, if x=0
Check whether f ′ f ′ f^(‘)f^{\prime}f′ is continuous on R R R\mathbb{R}R.
  1. a) Apply the Cauchy’s integral test to evaluate:
lim x → ∞ [ n + 1 n 2 + 1 2 + n + 2 n 2 + 2 2 + n + 3 n 2 + 3 2 + … + 1 n ] lim x → ∞   n + 1 n 2 + 1 2 + n + 2 n 2 + 2 2 + n + 3 n 2 + 3 2 + … + 1 n lim_(x rarr oo)[(n+1)/(n^(2)+1^(2))+(n+2)/(n^(2)+2^(2))+(n+3)/(n^(2)+3^(2))+dots+(1)/(n)]\lim _{x \rightarrow \infty}\left[\frac{n+1}{n^2+1^2}+\frac{n+2}{n^2+2^2}+\frac{n+3}{n^2+3^2}+\ldots+\frac{1}{n}\right]limx→∞[n+1n2+12+n+2n2+22+n+3n2+32+…+1n]
b) For x ∈ [ 0 , 2 ] x ∈ [ 0 , 2 ] x in[0,2]x \in[0,2]x∈[0,2] and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, define f n ( x ) = 3 x 2 + 2 x n f n ( x ) = 3 x 2 + 2 x n f_(n)(x)=3x^(2)+(2x)/(n)f_n(x)=3 x^2+\frac{2 x}{n}fn(x)=3x2+2xn. Find the limit function ‘ f f fff ‘ of the sequence ( f n ) n ∈ N f n n ∈ N (f_(n))_(n inN)\left(f_n\right)_{n \in \mathbb{N}}(fn)n∈N.Is f f fff continuous? Check if ∫ 0 2 f ( x ) d x ∫ 0 2   f ( x ) d x int_(0)^(2)f(x)dx\int_0^2 f(x) d x∫02f(x)dx and lim n → ∞ ∫ 0 2 f n ( x ) d x lim n → ∞   ∫ 0 2   f n ( x ) d x lim_(n rarr oo)int_(0)^(2)f_(n)(x)dx\lim _{n \rightarrow \infty} \int_0^2 f_n(x) d xlimn→∞∫02fn(x)dx are equal or not.
8. a) Find the radius of convergence of the series ∑ a n x n ∑ a n x n suma_(n)x^(n)\sum a_n x^n∑anxn where a n = n ! n n a n = n ! n n a_(n)=(n!)/(n^(n))a_n=\frac{n!}{n^n}an=n!nn.
b) f ( x ) = { x + 2 x 2 − 4 , when x ≠ − 2 − 1 , when x = − 2 f ( x ) = x + 2 x 2 − 4 ,      when x ≠ − 2 − 1 ,      when x = − 2 f(x)={[(x+2)/(x^(2)-4)”,”,” when “x!=-2],[-1″,”,” when “x=-2]:}f(x)= \begin{cases}\frac{x+2}{x^2-4}, & \text { when } x \neq-2 \\ -1, & \text { when } x=-2\end{cases}f(x)={x+2x2−4, when x≠−2−1, when x=−2
Check whether f f fff is uniformly continuous on [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1,1][−1,1] or not.
c) Show that
e − x < 1 − x + x 2 2 ! , x > 0 . e − x < 1 − x + x 2 2 ! , x > 0 . e^(-x) < 1-x+(x^(2))/(2!),x > 0.e^{-x}<1-x+\frac{x^2}{2!}, x>0 .e−x<1−x+x22!,x>0.
  1. a) Show that the sequence { f n } f n {f_(n)}\left\{f_n\right\}{fn} of functions, where
f n ( x ) = n x + n f n ( x ) = n x + n f_(n)(x)=(n)/(x+n)f_n(x)=\frac{n}{x+n}fn(x)=nx+n
is uniformly convergent in [ 0 , k ] [ 0 , k ] [0,k][0, k][0,k], where k > 0 k > 0 k > 0k>0k>0. Show further that { f n } f n {f_(n)}\left\{f_n\right\}{fn} is not uniformly convergent in [ 0 , ∞ [ [ 0 , ∞ [ [0,oo[[0, \infty[[0,∞[.
b) Evaluate ∫ 0 1 x 2 d x ∫ 0 1   x 2 d x int_(0)^(1)x^(2)dx\int_0^1 x^2 d x∫01x2dx using Riemann integration.
10. a) Find the value/s of x x xxx for which the series
∑ 1.3 .5 … . . ( 2 n − 1 ) 2.4 .6 … . .2 n ⋅ x n n ∑ 1.3 .5 … . . ( 2 n − 1 ) 2.4 .6 … . .2 n ⋅ x n n sum(1.3.5 dots..(2n-1))/(2.4.6 dots..2 n)*(x^(n))/(n)\sum \frac{1.3 .5 \ldots . .(2 n-1)}{2.4 .6 \ldots . .2 n} \cdot \frac{x^n}{n}∑1.3.5…..(2n−1)2.4.6…..2n⋅xnn
is convergent.
b) Give one example each for the following. Justify your choice of examples.
i) A sequence which is divergent.
ii) A set which is neither open nor closed.
iii) A compact set.
iv) A set which has no limit point.

Answer:

Question:-1

Which of the following statements are true or false? Give reasons for your answers in the form of a short proof or counter-example, whichever is appropriate:

i) Every infinite set is an open set.
ii) The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q.
iii) -1 is a limit point of the interval ] − 2 , 1 ] ] − 2 , 1 ] ]-2,1]]-2,1]]−2,1].
iv) The necessary condition for a function f f fff to be integrable is that it is continuous.
v) The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x)=|x-2|+|3-x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x=5x=5.

Answer:

i) Every infinite set is an open set.
False. In topology, a set is open if it contains an open neighborhood around each of its points (e.g., in the standard topology on R R R\mathbb{R}R, open intervals are open sets). Consider the infinite set Z Z Z\mathbb{Z}Z (the integers) in R R R\mathbb{R}R with the standard topology. The point 0 ∈ Z 0 ∈ Z 0inZ0 \in \mathbb{Z}0∈Z has no open interval ( a , b ) ( a , b ) (a,b)(a, b)(a,b) containing 0 0 000 that is entirely contained in Z Z Z\mathbb{Z}Z, since any such interval contains non-integer points. Thus, Z Z Z\mathbb{Z}Z is infinite but not open.
ii) The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is p → q p → q p rarr qp \rightarrow qp→q.
False. The negation of a statement is obtained by applying De Morgan’s laws. For p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q, the negation is:
∼ ( p ∧ ∼ q ) =∼ p ∨ ∼ ( ∼ q ) =∼ p ∨ q . ∼ ( p ∧ ∼ q ) =∼ p ∨ ∼ ( ∼ q ) =∼ p ∨ q . ∼(p^^∼q)=∼p vv∼(∼q)=∼p vv q.\sim (p \wedge \sim q) = \sim p \vee \sim (\sim q) = \sim p \vee q.∼(p∧∼q)=∼p∨∼(∼q)=∼p∨q.
Now, compare this to p → q p → q p rarr qp \rightarrow qp→q, which is equivalent to ∼ p ∨ q ∼ p ∨ q ∼p vv q\sim p \vee q∼p∨q. The two expressions are identical:
∼ p ∨ q =∼ p ∨ q . ∼ p ∨ q =∼ p ∨ q . ∼p vv q=∼p vv q.\sim p \vee q = \sim p \vee q.∼p∨q=∼p∨q.
Thus, the statement is True. The negation of p ∧ ∼ q p ∧ ∼ q p^^∼qp \wedge \sim qp∧∼q is indeed p → q p → q p rarr qp \rightarrow qp→q.
iii) -1 is a limit point of the interval ] − 2 , 1 ] ] − 2 , 1 ] ]-2,1]]-2,1]]−2,1].
True. A point x x xxx is a limit point of a set S S SSS if every neighborhood of x x xxx contains at least one point of S S SSS other than x x xxx. For the interval S = ( − 2 , 1 ] S = ( − 2 , 1 ] S=(-2,1]S = (-2, 1]S=(−2,1] in R R R\mathbb{R}R, consider x = − 1 x = − 1 x=-1x = -1x=−1. Any open interval ( − 1 − ϵ , − 1 + ϵ ) ( − 1 − ϵ , − 1 + ϵ ) (-1-epsilon,-1+epsilon)(-1 – \epsilon, -1 + \epsilon)(−1−ϵ,−1+ϵ) contains points in ( − 2 , − 1 ) ⊂ ( − 2 , 1 ] ( − 2 , − 1 ) ⊂ ( − 2 , 1 ] (-2,-1)sub(-2,1](-2, -1) \subset (-2, 1](−2,−1)⊂(−2,1], such as − 1 − ϵ / 2 − 1 − ϵ / 2 -1-epsilon//2-1 – \epsilon/2−1−ϵ/2, which are in S S SSS and distinct from − 1 − 1 -1-1−1. Thus, − 1 − 1 -1-1−1 is a limit point.
iv) The necessary condition for a function f f fff to be integrable is that it is continuous.
False. A function must be bounded and continuous almost everywhere (e.g., Riemann integrable) or measurable (e.g., Lebesgue integrable) to be integrable, but continuity is not strictly necessary. Consider the function f : [ 0 , 1 ] → R f : [ 0 , 1 ] → R f:[0,1]rarrRf: [0,1] \to \mathbb{R}f:[0,1]→R defined as f ( x ) = 1 f ( x ) = 1 f(x)=1f(x) = 1f(x)=1 if x x xxx is rational and f ( x ) = 0 f ( x ) = 0 f(x)=0f(x) = 0f(x)=0 if x x xxx is irrational. This function is discontinuous everywhere but is Riemann integrable on [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] with ∫ 0 1 f ( x ) d x = 0 ∫ 0 1   f ( x ) d x = 0 int_(0)^(1)f(x)dx=0\int_0^1 f(x) \, dx = 0∫01f(x)dx=0, since the set of rationals has measure zero.
v) The function f : R → R f : R → R f:RrarrRf: \mathbb{R} \rightarrow \mathbb{R}f:R→R defined by f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x) = |x-2| + |3-x|f(x)=|x−2|+|3−x| is differentiable at x = 5 x = 5 x=5x=5x=5.
True. To check differentiability, simplify f ( x ) = | x − 2 | + | 3 − x | f ( x ) = | x − 2 | + | 3 − x | f(x)=|x-2|+|3-x|f(x) = |x-2| + |3-x|f(x)=|x−2|+|3−x| at x = 5 x = 5 x=5x = 5x=5. Since x = 5 > 3 > 2 x = 5 > 3 > 2 x=5 > 3 > 2x = 5 > 3 > 2x=5>3>2, we have:
  • | x − 2 | = x − 2 | x − 2 | = x − 2 |x-2|=x-2|x-2| = x-2|x−2|=x−2,
  • | 3 − x | = − ( 3 − x ) = x − 3 | 3 − x | = − ( 3 − x ) = x − 3 |3-x|=-(3-x)=x-3|3-x| = -(3-x) = x-3|3−x|=−(3−x)=x−3.
Thus, for x ≥ 3 x ≥ 3 x >= 3x \geq 3x≥3,
f ( x ) = ( x − 2 ) + ( x − 3 ) = 2 x − 5. f ( x ) = ( x − 2 ) + ( x − 3 ) = 2 x − 5. f(x)=(x-2)+(x-3)=2x-5.f(x) = (x-2) + (x-3) = 2x – 5.f(x)=(x−2)+(x−3)=2x−5.
So, at x = 5 x = 5 x=5x = 5x=5, f ( x ) = 2 x − 5 f ( x ) = 2 x − 5 f(x)=2x-5f(x) = 2x – 5f(x)=2x−5, which is a linear function and differentiable everywhere with derivative f ′ ( x ) = 2 f ′ ( x ) = 2 f^(‘)(x)=2f'(x) = 2f′(x)=2. To confirm, check the derivative from both sides:
  • Right derivative: For x ≥ 5 x ≥ 5 x >= 5x \geq 5x≥5, f ( x ) = 2 x − 5 f ( x ) = 2 x − 5 f(x)=2x-5f(x) = 2x – 5f(x)=2x−5, so lim h → 0 + f ( 5 + h ) − f ( 5 ) h = lim h → 0 + ( 2 ( 5 + h ) − 5 ) − ( 2 ( 5 ) − 5 ) h = lim h → 0 + 2 h h = 2 lim h → 0 +   f ( 5 + h ) − f ( 5 ) h = lim h → 0 +   ( 2 ( 5 + h ) − 5 ) − ( 2 ( 5 ) − 5 ) h = lim h → 0 +   2 h h = 2 lim_(h rarr0^(+))(f(5+h)-f(5))/(h)=lim_(h rarr0^(+))((2(5+h)-5)-(2(5)-5))/(h)=lim_(h rarr0^(+))(2h)/(h)=2\lim_{h \to 0^+} \frac{f(5+h) – f(5)}{h} = \lim_{h \to 0^+} \frac{(2(5+h)-5) – (2(5)-5)}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2limh→0+f(5+h)−f(5)h=limh→0+(2(5+h)−5)−(2(5)−5)h=limh→0+2hh=2.
  • Left derivative: For 3 ≤ x ≤ 5 3 ≤ x ≤ 5 3 <= x <= 53 \leq x \leq 53≤x≤5, f ( x ) = 2 x − 5 f ( x ) = 2 x − 5 f(x)=2x-5f(x) = 2x – 5f(x)=2x−5 (since | x − 2 | = x − 2 | x − 2 | = x − 2 |x-2|=x-2|x-2| = x-2|x−2|=x−2, | 3 − x | = − ( 3 − x ) = x − 3 | 3 − x | = − ( 3 − x ) = x − 3 |3-x|=-(3-x)=x-3|3-x| = -(3-x) = x-3|3−x|=−(3−x)=x−3), so the left derivative is also 2 2 222. For x < 3 x < 3 x < 3x < 3x<3, compute f ( x ) = | x − 2 | + | 3 − x | = ( x − 2 ) + ( 3 − x ) = 1 f ( x ) = | x − 2 | + | 3 − x | = ( x − 2 ) + ( 3 − x ) = 1 f(x)=|x-2|+|3-x|=(x-2)+(3-x)=1f(x) = |x-2| + |3-x| = (x-2) + (3-x) = 1f(x)=|x−2|+|3−x|=(x−2)+(3−x)=1, which is constant, so the derivative is 0 0 000. However, since x = 5 x = 5 x=5x = 5x=5 is in the region x ≥ 3 x ≥ 3 x >= 3x \geq 3x≥3, we only need continuity of the derivative at x = 5 x = 5 x=5x = 5x=5, which holds since f ( x ) = 2 x − 5 f ( x ) = 2 x − 5 f(x)=2x-5f(x) = 2x – 5f(x)=2x−5 applies in a neighborhood of 5 5 555.
Since the left and right derivatives exist and are equal ( 2 2 222), f f fff is differentiable at x = 5 x = 5 x=5x = 5x=5.

Question:-2(a)

Test the following series for convergence:

(i) ∑ n = 1 ∞ n x n − 1 , x > 0 ∑ n = 1 ∞   n x n − 1 , x > 0 sum_(n=1)^(oo)nx^(n-1),x > 0\sum_{n=1}^{\infty} n x^{n-1}, x > 0∑n=1∞nxn−1,x>0.
(ii) ∑ n = 1 ∞ [ n 4 + 9 − n 4 − 9 ] ∑ n = 1 ∞   n 4 + 9 − n 4 − 9 sum_(n=1)^(oo)[sqrt(n^(4)+9)-sqrt(n^(4)-9)]\sum_{n=1}^{\infty} \left[ \sqrt{n^4 + 9} – \sqrt{n^4 – 9} \right]∑n=1∞[n4+9−n4−9].

Answer:

(i) ∑ n = 1 ∞ n x n − 1 , x > 0 ∑ n = 1 ∞   n x n − 1 , x > 0 sum_(n=1)^(oo)nx^(n-1),x > 0\sum_{n=1}^{\infty} n x^{n-1}, x > 0∑n=1∞nxn−1,x>0

Rewrite the series to understand its form:
∑ n = 1 ∞ n x n − 1 = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ ∑ n = 1 ∞   n x n − 1 = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ sum_(n=1)^(oo)nx^(n-1)=1+2x+3x^(2)+4x^(3)+cdots\sum_{n=1}^{\infty} n x^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \cdots∑n=1∞nxn−1=1+2x+3x2+4x3+⋯
This resembles the derivative of a geometric series. Recall the geometric series:
∑ n = 0 ∞ x n = 1 1 − x , | x | < 1. ∑ n = 0 ∞   x n = 1 1 − x , | x | < 1. sum_(n=0)^(oo)x^(n)=(1)/(1-x),quad|x| < 1.\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, \quad |x| < 1.∑n=0∞xn=11−x,|x|<1.
Differentiate both sides with respect to x x xxx:
∑ n = 1 ∞ n x n − 1 = d d x ( ∑ n = 0 ∞ x n ) = d d x ( 1 1 − x ) = 1 ( 1 − x ) 2 . ∑ n = 1 ∞   n x n − 1 = d d x ∑ n = 0 ∞   x n = d d x 1 1 − x = 1 ( 1 − x ) 2 . sum_(n=1)^(oo)nx^(n-1)=(d)/(dx)(sum_(n=0)^(oo)x^(n))=(d)/(dx)((1)/(1-x))=(1)/((1-x)^(2)).\sum_{n=1}^{\infty} n x^{n-1} = \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right) = \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2}.∑n=1∞nxn−1=ddx(∑n=0∞xn)=ddx(11−x)=1(1−x)2.
This holds for | x | < 1 | x | < 1 |x| < 1|x| < 1|x|<1, so the series converges to 1 ( 1 − x ) 2 1 ( 1 − x ) 2 (1)/((1-x)^(2))\frac{1}{(1-x)^2}1(1−x)2 when 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1 (since x > 0 x > 0 x > 0x > 0x>0).
Now, test the boundaries and beyond:
  • For x ≥ 1 x ≥ 1 x >= 1x \geq 1x≥1: Consider the terms n x n − 1 n x n − 1 nx^(n-1)n x^{n-1}nxn−1. If x = 1 x = 1 x=1x = 1x=1, the series becomes ∑ n = 1 ∞ n = 1 + 2 + 3 + ⋯ ∑ n = 1 ∞   n = 1 + 2 + 3 + ⋯ sum_(n=1)^(oo)n=1+2+3+cdots\sum_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots∑n=1∞n=1+2+3+⋯, which diverges (the partial sums grow without bound). If x > 1 x > 1 x > 1x > 1x>1, the terms n x n − 1 = n x n / x n x n − 1 = n x n / x nx^(n-1)=nx^(n)//xn x^{n-1} = n x^n / xnxn−1=nxn/x grow exponentially since x n x n x^(n)x^nxn dominates, and the series diverges (e.g., terms do not approach 0, a necessary condition for convergence).
  • For 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1: As derived, the series converges to 1 ( 1 − x ) 2 1 ( 1 − x ) 2 (1)/((1-x)^(2))\frac{1}{(1-x)^2}1(1−x)2.
  • At x = 1 x = 1 x=1x = 1x=1: The series ∑ n ∑ n sum n\sum n∑n diverges, as noted.
Thus, the series converges for 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1 and diverges for x ≥ 1 x ≥ 1 x >= 1x \geq 1x≥1.
Conclusion for (i): The series ∑ n = 1 ∞ n x n − 1 ∑ n = 1 ∞   n x n − 1 sum_(n=1)^(oo)nx^(n-1)\sum_{n=1}^{\infty} n x^{n-1}∑n=1∞nxn−1 converges for 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1 to 1 ( 1 − x ) 2 1 ( 1 − x ) 2 (1)/((1-x)^(2))\frac{1}{(1-x)^2}1(1−x)2 and diverges for x ≥ 1 x ≥ 1 x >= 1x \geq 1x≥1.

(ii) ∑ n = 1 ∞ ( n 4 + 9 − n 4 − 9 ) ∑ n = 1 ∞   n 4 + 9 − n 4 − 9 sum_(n=1)^(oo)(sqrt(n^(4)+9)-sqrt(n^(4)-9))\sum_{n=1}^{\infty} \left( \sqrt{n^4 + 9} – \sqrt{n^4 – 9} \right)∑n=1∞(n4+9−n4−9)

To determine convergence, analyze the general term a n = n 4 + 9 − n 4 − 9 a n = n 4 + 9 − n 4 − 9 a_(n)=sqrt(n^(4)+9)-sqrt(n^(4)-9)a_n = \sqrt{n^4 + 9} – \sqrt{n^4 – 9}an=n4+9−n4−9. Since the terms are positive, we need to check if the sum ∑ a n ∑ a n suma_(n)\sum a_n∑an is finite.
Step 1: Approximate the general term for large n n nnn.
For large n n nnn, n 4 n 4 n^(4)n^4n4 dominates, so approximate:
n 4 + 9 ≈ n 2 + 9 2 n 2 , n 4 − 9 ≈ n 2 − 9 2 n 2 . n 4 + 9 ≈ n 2 + 9 2 n 2 , n 4 − 9 ≈ n 2 − 9 2 n 2 . sqrt(n^(4)+9)~~n^(2)+(9)/(2n^(2)),quadsqrt(n^(4)-9)~~n^(2)-(9)/(2n^(2)).\sqrt{n^4 + 9} \approx n^2 + \frac{9}{2n^2}, \quad \sqrt{n^4 – 9} \approx n^2 – \frac{9}{2n^2}.n4+9≈n2+92n2,n4−9≈n2−92n2.
More precisely, use the binomial approximation for 1 + x ≈ 1 + x 2 1 + x ≈ 1 + x 2 sqrt(1+x)~~1+(x)/(2)\sqrt{1 + x} \approx 1 + \frac{x}{2}1+x≈1+x2 when x x xxx is small:
n 4 + 9 = n 2 1 + 9 n 4 ≈ n 2 ( 1 + 9 2 n 4 ) = n 2 + 9 2 n 2 , n 4 + 9 = n 2 1 + 9 n 4 ≈ n 2 1 + 9 2 n 4 = n 2 + 9 2 n 2 , sqrt(n^(4)+9)=n^(2)sqrt(1+(9)/(n^(4)))~~n^(2)(1+(9)/(2n^(4)))=n^(2)+(9)/(2n^(2)),\sqrt{n^4 + 9} = n^2 \sqrt{1 + \frac{9}{n^4}} \approx n^2 \left( 1 + \frac{9}{2n^4} \right) = n^2 + \frac{9}{2n^2},n4+9=n21+9n4≈n2(1+92n4)=n2+92n2,
n 4 − 9 = n 2 1 − 9 n 4 ≈ n 2 ( 1 − 9 2 n 4 ) = n 2 − 9 2 n 2 . n 4 − 9 = n 2 1 − 9 n 4 ≈ n 2 1 − 9 2 n 4 = n 2 − 9 2 n 2 . sqrt(n^(4)-9)=n^(2)sqrt(1-(9)/(n^(4)))~~n^(2)(1-(9)/(2n^(4)))=n^(2)-(9)/(2n^(2)).\sqrt{n^4 – 9} = n^2 \sqrt{1 – \frac{9}{n^4}} \approx n^2 \left( 1 – \frac{9}{2n^4} \right) = n^2 – \frac{9}{2n^2}.n4−9=n21−9n4≈n2(1−92n4)=n2−92n2.
Thus,
a n ≈ ( n 2 + 9 2 n 2 ) − ( n 2 − 9 2 n 2 ) = 9 2 n 2 + 9 2 n 2 = 9 n 2 . a n ≈ n 2 + 9 2 n 2 − n 2 − 9 2 n 2 = 9 2 n 2 + 9 2 n 2 = 9 n 2 . a_(n)~~(n^(2)+(9)/(2n^(2)))-(n^(2)-(9)/(2n^(2)))=(9)/(2n^(2))+(9)/(2n^(2))=(9)/(n^(2)).a_n \approx \left( n^2 + \frac{9}{2n^2} \right) – \left( n^2 – \frac{9}{2n^2} \right) = \frac{9}{2n^2} + \frac{9}{2n^2} = \frac{9}{n^2}.an≈(n2+92n2)−(n2−92n2)=92n2+92n2=9n2.
Step 2: Rationalize the expression for exact behavior.
To confirm, rationalize the expression:
a n = n 4 + 9 − n 4 − 9 = ( n 4 + 9 ) − ( n 4 − 9 ) n 4 + 9 + n 4 − 9 = 18 n 4 + 9 + n 4 − 9 . a n = n 4 + 9 − n 4 − 9 = ( n 4 + 9 ) − ( n 4 − 9 ) n 4 + 9 + n 4 − 9 = 18 n 4 + 9 + n 4 − 9 . a_(n)=sqrt(n^(4)+9)-sqrt(n^(4)-9)=((n^(4)+9)-(n^(4)-9))/(sqrt(n^(4)+9)+sqrt(n^(4)-9))=(18)/(sqrt(n^(4)+9)+sqrt(n^(4)-9)).a_n = \sqrt{n^4 + 9} – \sqrt{n^4 – 9} = \frac{(n^4 + 9) – (n^4 – 9)}{\sqrt{n^4 + 9} + \sqrt{n^4 – 9}} = \frac{18}{\sqrt{n^4 + 9} + \sqrt{n^4 – 9}}.an=n4+9−n4−9=(n4+9)−(n4−9)n4+9+n4−9=18n4+9+n4−9.
For large n n nnn,
n 4 + 9 + n 4 − 9 ≈ n 2 + n 2 = 2 n 2 , n 4 + 9 + n 4 − 9 ≈ n 2 + n 2 = 2 n 2 , sqrt(n^(4)+9)+sqrt(n^(4)-9)~~n^(2)+n^(2)=2n^(2),\sqrt{n^4 + 9} + \sqrt{n^4 – 9} \approx n^2 + n^2 = 2n^2,n4+9+n4−9≈n2+n2=2n2,
so,
a n ≈ 18 2 n 2 = 9 n 2 . a n ≈ 18 2 n 2 = 9 n 2 . a_(n)~~(18)/(2n^(2))=(9)/(n^(2)).a_n \approx \frac{18}{2n^2} = \frac{9}{n^2}.an≈182n2=9n2.
Step 3: Apply the comparison test.
Since a n ≈ 9 n 2 a n ≈ 9 n 2 a_(n)~~(9)/(n^(2))a_n \approx \frac{9}{n^2}an≈9n2, compare with the series ∑ 9 n 2 = 9 ∑ 1 n 2 ∑ 9 n 2 = 9 ∑ 1 n 2 sum(9)/(n^(2))=9sum(1)/(n^(2))\sum \frac{9}{n^2} = 9 \sum \frac{1}{n^2}∑9n2=9∑1n2. The p-series ∑ 1 n 2 ∑ 1 n 2 sum(1)/(n^(2))\sum \frac{1}{n^2}∑1n2 converges ( p = 2 > 1 p = 2 > 1 p=2 > 1p = 2 > 1p=2>1), so ∑ 9 n 2 ∑ 9 n 2 sum(9)/(n^(2))\sum \frac{9}{n^2}∑9n2 converges.
To make the comparison rigorous, estimate the denominator:
n 4 + 9 ≤ n 2 + 9 = n 2 + 3 , n 4 − 9 ≤ n 2 , n 4 + 9 ≤ n 2 + 9 = n 2 + 3 , n 4 − 9 ≤ n 2 , sqrt(n^(4)+9) <= n^(2)+sqrt9=n^(2)+3,quadsqrt(n^(4)-9) <= n^(2),\sqrt{n^4 + 9} \leq n^2 + \sqrt{9} = n^2 + 3, \quad \sqrt{n^4 – 9} \leq n^2,n4+9≤n2+9=n2+3,n4−9≤n2,
n 4 + 9 + n 4 − 9 ≤ ( n 2 + 3 ) + n 2 = 2 n 2 + 3. n 4 + 9 + n 4 − 9 ≤ ( n 2 + 3 ) + n 2 = 2 n 2 + 3. sqrt(n^(4)+9)+sqrt(n^(4)-9) <= (n^(2)+3)+n^(2)=2n^(2)+3.\sqrt{n^4 + 9} + \sqrt{n^4 – 9} \leq (n^2 + 3) + n^2 = 2n^2 + 3.n4+9+n4−9≤(n2+3)+n2=2n2+3.
Thus,
a n = 18 n 4 + 9 + n 4 − 9 ≥ 18 2 n 2 + 3 . a n = 18 n 4 + 9 + n 4 − 9 ≥ 18 2 n 2 + 3 . a_(n)=(18)/(sqrt(n^(4)+9)+sqrt(n^(4)-9)) >= (18)/(2n^(2)+3).a_n = \frac{18}{\sqrt{n^4 + 9} + \sqrt{n^4 – 9}} \geq \frac{18}{2n^2 + 3}.an=18n4+9+n4−9≥182n2+3.
Also, since n 4 + 9 ≥ n 2 n 4 + 9 ≥ n 2 sqrt(n^(4)+9) >= n^(2)\sqrt{n^4 + 9} \geq n^2n4+9≥n2, n 4 − 9 ≥ n 2 − 9 = n 2 − 3 n 4 − 9 ≥ n 2 − 9 = n 2 − 3 sqrt(n^(4)-9) >= n^(2)-sqrt9=n^(2)-3\sqrt{n^4 – 9} \geq n^2 – \sqrt{9} = n^2 – 3n4−9≥n2−9=n2−3,
n 4 + 9 + n 4 − 9 ≥ n 2 + ( n 2 − 3 ) = 2 n 2 − 3 , n 4 + 9 + n 4 − 9 ≥ n 2 + ( n 2 − 3 ) = 2 n 2 − 3 , sqrt(n^(4)+9)+sqrt(n^(4)-9) >= n^(2)+(n^(2)-3)=2n^(2)-3,\sqrt{n^4 + 9} + \sqrt{n^4 – 9} \geq n^2 + (n^2 – 3) = 2n^2 – 3,n4+9+n4−9≥n2+(n2−3)=2n2−3,
a n ≤ 18 2 n 2 − 3 . a n ≤ 18 2 n 2 − 3 . a_(n) <= (18)/(2n^(2)-3).a_n \leq \frac{18}{2n^2 – 3}.an≤182n2−3.
For large n n nnn,
18 2 n 2 + 3 ≈ 18 2 n 2 = 9 n 2 , 18 2 n 2 − 3 ≈ 18 2 n 2 = 9 n 2 . 18 2 n 2 + 3 ≈ 18 2 n 2 = 9 n 2 , 18 2 n 2 − 3 ≈ 18 2 n 2 = 9 n 2 . (18)/(2n^(2)+3)~~(18)/(2n^(2))=(9)/(n^(2)),quad(18)/(2n^(2)-3)~~(18)/(2n^(2))=(9)/(n^(2)).\frac{18}{2n^2 + 3} \approx \frac{18}{2n^2} = \frac{9}{n^2}, \quad \frac{18}{2n^2 – 3} \approx \frac{18}{2n^2} = \frac{9}{n^2}.182n2+3≈182n2=9n2,182n2−3≈182n2=9n2.
Thus, 18 2 n 2 + 3 ≤ a n ≤ 18 2 n 2 − 3 18 2 n 2 + 3 ≤ a n ≤ 18 2 n 2 − 3 (18)/(2n^(2)+3) <= a_(n) <= (18)/(2n^(2)-3)\frac{18}{2n^2 + 3} \leq a_n \leq \frac{18}{2n^2 – 3}182n2+3≤an≤182n2−3, and both bounding series behave like 9 n 2 9 n 2 (9)/(n^(2))\frac{9}{n^2}9n2. Since ∑ 18 2 n 2 + 3 ∑ 18 2 n 2 + 3 sum(18)/(2n^(2)+3)\sum \frac{18}{2n^2 + 3}∑182n2+3 and ∑ 18 2 n 2 − 3 ∑ 18 2 n 2 − 3 sum(18)/(2n^(2)-3)\sum \frac{18}{2n^2 – 3}∑182n2−3 are comparable to ∑ 9 n 2 ∑ 9 n 2 sum(9)/(n^(2))\sum \frac{9}{n^2}∑9n2 (which converges), use the limit comparison test with b n = 9 n 2 b n = 9 n 2 b_(n)=(9)/(n^(2))b_n = \frac{9}{n^2}bn=9n2:
lim n → ∞ a n b n = lim n → ∞ 18 n 4 + 9 + n 4 − 9 9 n 2 = lim n → ∞ 18 n 2 9 ( n 4 + 9 + n 4 − 9 ) = 18 n 2 9 ⋅ 2 n 2 = 1. lim n → ∞   a n b n = lim n → ∞   18 n 4 + 9 + n 4 − 9 9 n 2 = lim n → ∞   18 n 2 9 ( n 4 + 9 + n 4 − 9 ) = 18 n 2 9 ⋅ 2 n 2 = 1. lim_(n rarr oo)(a_(n))/(b_(n))=lim_(n rarr oo)((18)/(sqrt(n^(4)+9)+sqrt(n^(4)-9)))/((9)/(n^(2)))=lim_(n rarr oo)(18n^(2))/(9(sqrt(n^(4)+9)+sqrt(n^(4)-9)))=(18n^(2))/(9*2n^(2))=1.\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{18}{\sqrt{n^4 + 9} + \sqrt{n^4 – 9}}}{\frac{9}{n^2}} = \lim_{n \to \infty} \frac{18 n^2}{9 (\sqrt{n^4 + 9} + \sqrt{n^4 – 9})} = \frac{18 n^2}{9 \cdot 2n^2} = 1.limn→∞anbn=limn→∞18n4+9+n4−99n2=limn→∞18n29(n4+9+n4−9)=18n29⋅2n2=1.
Since the limit is finite and positive, and ∑ b n = ∑ 9 n 2 ∑ b n = ∑ 9 n 2 sumb_(n)=sum(9)/(n^(2))\sum b_n = \sum \frac{9}{n^2}∑bn=∑9n2 converges, ∑ a n ∑ a n suma_(n)\sum a_n∑an converges.
Conclusion for (ii): The series ∑ n = 1 ∞ ( n 4 + 9 − n 4 − 9 ) ∑ n = 1 ∞   n 4 + 9 − n 4 − 9 sum_(n=1)^(oo)(sqrt(n^(4)+9)-sqrt(n^(4)-9))\sum_{n=1}^{\infty} \left( \sqrt{n^4 + 9} – \sqrt{n^4 – 9} \right)∑n=1∞(n4+9−n4−9) converges.

Final Answer

  • (i): Converges for 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1 to 1 ( 1 − x ) 2 1 ( 1 − x ) 2 (1)/((1-x)^(2))\frac{1}{(1-x)^2}1(1−x)2, diverges for x ≥ 1 x ≥ 1 x >= 1x \geq 1x≥1.
  • (ii): Converges.
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