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Q. 36

Expert-verified
Found in: Page 625

Calculus

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.

See the step by step solution

Step 1. Given information

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.

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

The integral diverges.

So, the series is divergent.