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.
Using the comparison test, we conclude that sumv_(n)\sum v_n is convergent when p > 1p>1.
Step 3: Case 1 – Absolute Convergence for P > 1P > 1:
sumu_(n)\sum u_n is an alternating series and sum|u_(n)|\sum\left|u_n\right| is convergent. Therefore, sumu_(n)\sum u_n is absolutely convergent.
Step 4: Case 2 – Conditional Convergence for 0 < p <= 10<p \leq 1:
When 0 < p <= 10<p \leq 1, the sequence {v_(n)}\{v_n\} is a monotone decreasing sequence of positive real numbers with limv_(n)=0\lim v_n = 0. By Leibniz’s test, sum(-1)^(n+1)v_(n)\sum (-1)^{n+1} v_n, which is equivalent to sumu_(n)\sum u_n, is convergent.
Since sum|u_(n)|\sum\left|u_n\right| is divergent in this case, sumu_(n)\sum u_n is conditionally convergent.
Step 5: Decomposition of u_(n)u_n into P_(n)P_n and q_(n)q_n:
Let Sigmau_(n)\Sigma u_n be a series of positive real numbers. We can decompose it into two new series:
In summary, the given series sumu_(n)\sum u_n is absolutely convergent when P > 1P > 1, and it is conditionally convergent when 0 < p <= 10<p \leq 1. This analysis provides a comprehensive understanding of the convergence behavior of the series for different values of pp in the specified range.