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Q. 36

Expert-verifiedFound in: Page 625

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use either the divergence test or the integral test to determine whether the series in Exercises 32–43 converge or diverge. Explain why the series meets the hypotheses of the test you select.

36. $\sum _{k=1}^{\infty}\frac{k}{{k}^{2}+3}$

The series is divergent.

We have been given the series $\sum _{k=1}^{\infty}\frac{k}{{k}^{2}+3}$

We have to determine whether the series converge or diverge.

Consider function $f\left(x\right)=\frac{x}{3+{x}^{2}}$

The function is continuous, decreasing, with positive terms.

All the conditions of integral test are fulfilled.

So, integral test is applicable.

Consider the integral localid="1649088344379" ${\int}_{x=1}^{\infty}f\left(x\right)dx={\int}_{x=1}^{\infty}\frac{x}{3+{x}^{2}}dx$

localid="1649088398964" ${\int}_{x=1}^{\infty}f\left(x\right)dx\phantom{\rule{0ex}{0ex}}=\underset{k\to \infty}{\mathrm{lim}}{\int}_{x=1}^{k}\frac{x}{3+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{k\to \infty}{\mathrm{lim}}{\int}_{u=4}^{{k}^{2}+3}\frac{du}{u}(Put3+{x}^{2}=u\Rightarrow 2xdx=du)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{k\to \infty}{\mathrm{lim}}{\left[\mathrm{ln}\left|u\right|\right]}_{4}^{{k}^{2}+3}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{k\to \infty}{\mathrm{lim}}\left[\mathrm{ln}\left|{k}^{2}+3\right|-\mathrm{ln}\left|4\right|\right]\phantom{\rule{0ex}{0ex}}=\infty $

The integral diverges.

So, the series is divergent.

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