/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Free solutions & answers for Calculus Early Transcendentals Chapter 11 - (Page 32) [step by step] | 91Ó°ÊÓ

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Problem 31

Test the series for convergence or divergence. $$\sum_{k=1}^{\infty} \frac{5^{k}}{3^{k}+4^{k}}$$

Problem 31

Determine whether the sequence converges or diverges. If it converges, find the limit. $$ \left\\{\frac{e^{n}+e^{-n}}{e^{2 n}-1}\right\\} $$

Problem 31

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. \(f(x)=e^{x}+e^{2 x}\)

Problem 31

\(3-32\) Determine whether the series converges or diverges. $$\sum_{n=1}^{\infty} \sin \left(\frac{1}{n}\right)$$

Problem 31

For which of the following series is the Ratio Test inconclusive (that is, it fails to give a definite answer)? $$\begin{array}{ll}{\text { (a) } \sum_{n=1}^{\infty} \frac{1}{n^{3}}} & {\text { (b) } \sum_{n=1}^{\infty} \frac{n}{2^{n}}} \\ {\text { (c) } \sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{\sqrt{n}}} & {\text { (d) } \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^{2}}}\end{array}$$

Problem 31

Is the 50 th partial sum \(s_{50}\) of the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1} / n\) an overestimate or an underestimate of the total sum? Explain.

Problem 31

\(21-34\) Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \arctan n$$

Problem 32

\(21-34\) Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty}\left(\frac{3}{5^{n}}+\frac{2}{n}\right)$$

Problem 32

\(3-32\) Determine whether the series converges or diverges. $$\sum_{n=1}^{\infty} \frac{1}{n^{1+1 / n}}$$

Problem 32

Determine whether the sequence converges or diverges. If it converges, find the limit. $$ \left\\{\frac{\ln n}{\ln 2 n}\right\\} $$

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