Solved: In Exercises 1136, determine the convergence or divergence of the series. n1

Chapter 9, Problem 36

(choose chapter or problem)

In Exercises 11-36, determine the convergence or divergence of the series.

\(\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{e^{n}+e^{-n}}=\sum_{n=1}^{\infty}(-1)^{n+1} {sech} n\)

Text Transcription:

sum_n=1 ^infty 2(-1)^n+1 / e^n + e^-n = sum_n=1 ^infty (-1)^n+1 sech n

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