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Question:-1(d)

\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)
Find the range of \(p(>0)\) for which the series: \(\frac{1}{(1+a)^p}-\frac{1}{(2+a)^p}+\frac{1}{(3+a)^p}-\ldots, a>0\), is (i) absolutely convergent and (ii) conditionally convergent.
Expert Answer
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Introduction:
We are tasked with finding the range of p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0) for which the series given by:
1 ( 1 + a ) p 1 ( 2 + a ) p + 1 ( 3 + a ) p , where a > 0 1 ( 1 + a ) p 1 ( 2 + a ) p + 1 ( 3 + a ) p ,  where  a > 0 (1)/((1+a)^(p))-(1)/((2+a)^(p))+(1)/((3+a)^(p))-dots,” where “a > 0\frac{1}{(1+a)^p}-\frac{1}{(2+a)^p}+\frac{1}{(3+a)^p}-\ldots, \text{ where } a>01(1+a)p1(2+a)p+1(3+a)p, where a>0
converges absolutely and conditionally. To do this, we will use various convergence tests and mathematical techniques.
Step 1: Definition of v n v n v_(n)v_nvn and ω n ω n omega _(n)\omega_nωn:
Let u n u n sumu_(n)\sum u_nun be the given series, and define a new series v n v n v_(n)v_nvn as follows:
v n = 1 ( n + a ) p v n = 1 ( n + a ) p v_(n)=(1)/((n+a)^(p))v_n = \frac{1}{(n+a)^p}vn=1(n+a)p
Additionally, define ω n ω n omega _(n)\omega_nωn as:
ω n = 1 n p ω n = 1 n p omega _(n)=(1)/(n^(p))\omega_n = \frac{1}{n^p}ωn=1np
Step 2: Checking Convergence of v n v n sumv_(n)\sum v_nvn for p > 1 p > 1 p > 1p>1p>1:
By comparing v n v n v_(n)v_nvn with ω n ω n omega _(n)\omega_nωn, we find that:
lim n v n ω n = 1 lim n v n ω n = 1 lim_(n rarr oo)(v_(n))/(omega _(n))=1\lim_{n\to\infty} \frac{v_n}{\omega_n} = 1limnvnωn=1
Using the comparison test, we conclude that v n v n sumv_(n)\sum v_nvn is convergent when p > 1 p > 1 p > 1p>1p>1.
Step 3: Case 1 – Absolute Convergence for P > 1 P > 1 P > 1P > 1P>1:
u n u n sumu_(n)\sum u_nun is an alternating series and | u n | u n sum|u_(n)|\sum\left|u_n\right||un| is convergent. Therefore, u n u n sumu_(n)\sum u_nun is absolutely convergent.
Step 4: Case 2 – Conditional Convergence for 0 < p 1 0 < p 1 0 < p <= 10<p \leq 10<p1:
When 0 < p 1 0 < p 1 0 < p <= 10<p \leq 10<p1, the sequence { v n } { v n } {v_(n)}\{v_n\}{vn} is a monotone decreasing sequence of positive real numbers with lim v n = 0 lim v n = 0 limv_(n)=0\lim v_n = 0limvn=0. By Leibniz’s test, ( 1 ) n + 1 v n ( 1 ) n + 1 v n sum(-1)^(n+1)v_(n)\sum (-1)^{n+1} v_n(1)n+1vn, which is equivalent to u n u n sumu_(n)\sum u_nun, is convergent.
Since | u n | u n sum|u_(n)|\sum\left|u_n\right||un| is divergent in this case, u n u n sumu_(n)\sum u_nun is conditionally convergent.
Step 5: Decomposition of u n u n u_(n)u_nun into P n P n P_(n)P_nPn and q n q n q_(n)q_nqn:
Let Σ u n Σ u n Sigmau_(n)\Sigma u_nΣun be a series of positive real numbers. We can decompose it into two new series:
P n = { u n if u n > 0 0 if u n 0 q n = { 0 if u n 0 u n if u n < 0 P n = u n if  u n > 0 0 if  u n 0 q n = 0 if  u n 0 u n if  u n < 0 {:[P_(n)={[u_(n),”if “u_(n) > 0],[0,”if “u_(n) <= 0]:}],[q_(n)={[0,”if “u_(n) >= 0],[u_(n),”if “u_(n) < 0]:}]:}\begin{align*} P_n &= \begin{cases} u_n & \text{if } u_n > 0 \\ 0 & \text{if } u_n \leq 0 \end{cases} \\ q_n &= \begin{cases} 0 & \text{if } u_n \geq 0 \\ u_n & \text{if } u_n < 0 \end{cases} \end{align*}Pn={unif un>00if un0qn={0if un0unif un<0
Conclusion:
In summary, the given series u n u n sumu_(n)\sum u_nun is absolutely convergent when P > 1 P > 1 P > 1P > 1P>1, and it is conditionally convergent when 0 < p 1 0 < p 1 0 < p <= 10<p \leq 10<p1. This analysis provides a comprehensive understanding of the convergence behavior of the series for different values of p p ppp in the specified range.
Verified Answer
5/5
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