Here are some telescoping series problems: a. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ b. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series. c. Sum the series \(\sum_{n=1}^{\infty}\left[\tan ^{-1} n-\tan ^{-1}(n+1)\right]\) by first writing the \(N\) th partial sum and then computing \(\lim _{N \rightarrow \infty} s_{N}\)

Short Answer

Expert verified
a) The identity is verified by writing out the first few terms and observing the cancelations. b) The sum of the series is 1. c) The sum of the series is \(\frac{\pi}{4}\).

Step by step solution

01

Part (a) Verification of the Identity

Write out a few terms of the sum to verify the identity. Cancellation of terms will occur, and the proof of the identity relies on these cancellations.
02

Part (b) Find the nth Partial Sum

To find the nth partial sum (denoted \(S_n\)), write the sum and withdraw terms until a pattern appears. From there, generalize \(S_n\) in terms of n.
03

Part (b) Find the Sum of the Series

To find the sum of the series, compute \(\lim_{n \to \infty} S_n\). The sum of the series, if it exists, is this limit.
04

Part (c) Rewrite the Series

Write the series as the partial sum, \(S_N\), then compute \(\lim_{N \rightarrow \infty} S_N\). The sum of this series, if it exists, is this limit.

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