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Q. 24

Expert-verifiedFound in: Page 625

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.

$\sum _{k=1}^{\mathrm{\infty}}\u200a\frac{k}{{e}^{k}}$

Ans: The series $\sum _{k=1}^{\mathrm{\infty}}\u200a\frac{k}{{e}^{k}}$is convergent.

given,

$\sum _{k=1}^{\mathrm{\infty}}\u200a\frac{k}{{e}^{k}}$

Consider function $f\left(x\right)=x{e}^{-x}$.

The first derivative of the function $f\left(x\right)=x{e}^{-x}$ is:

$\begin{array}{r}{f}^{\mathrm{\prime}}\left(x\right)={e}^{-x}-x{e}^{-x}\\ =(1-x){e}^{-x}\end{array}$

The derivative is negative for all $x>1$. Therefore the function is decreasing.

The function $f\left(x\right)=x{e}^{-x}$ is continuous, decreasing, with positive terms. Therefore, all the conditions of the integral test are fulfilled. So, the integral test is applicable.

Therefore,

$\begin{array}{r}{\int}_{x=1}^{\mathrm{\infty}}\u200ax{e}^{-x}dx=\underset{k\to \mathrm{\infty}}{lim}\u200a{\int}_{x=1}^{k}\u200ax{e}^{-x}dx\\ =\underset{k\to \mathrm{\infty}}{lim}\u200a{\left[-x{e}^{-x}-\left({e}^{-x}\right)\right]}_{x=1}^{k}\text{(Integrate)}\\ =\underset{k\to \mathrm{\infty}}{lim}\u200a{\left[-x{e}^{-x}-{e}^{-x}\right]}_{x=1}^{k}\\ =\underset{k\to \mathrm{\infty}}{lim}\u200a\left[-k{e}^{-k}-{e}^{-k}+2{e}^{-1}\right]\text{(Substitution)}\\ =2{e}^{-1}\text{(Take limit)}\end{array}$

The integral converges. Therefore, the series $\sum _{k=1}^{\mathrm{\infty}}\u200a\frac{k}{{e}^{k}}$ is convergent.

Hence, by integral test, the series $\sum _{k=1}^{\mathrm{\infty}}\u200a\frac{k}{{e}^{k}}$ is convergent.

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