Q. 56 - Calculus

Question

Use any convergence test from Sections 7.4–7.6 to determine whether the series in Exercises 41–59 converge or diverge. Explain why each series that meets the hypotheses of the test you select does so. 

$\sum _{k=1}^{\infty }\frac{e^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}}}{k^2}$

Step-by-Step Solution

Answer

The given series converges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{e^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}}}{k^2}.$

Step 2. Determine whether the series converges or diverges.

To determine whether the series converges or diverges we will use the integral test since the series has positive terms and continuous that meet the hypothesis of the test. 

Let's check the convergence of ${\int }_1^{\infty }\frac{e^{{\displaystyle \frac{1}{k}}}}{k^2}dk.$

Let $u=\frac{1}{k}, \mathrm{so} \mathrm{the} \mathrm{derivation} \mathrm{is} du=-\frac{1}{k^2}dk.$

So,

$$\begin{gathered} {\int }_1^{\infty }-e^{u}du \\ ={\left[-e^u\right]}_1^{\infty } \\ ={\left[-e^{\frac{1}{k}}\right]}_1^{\infty } \\ =-1-\left(-e\right) \\ =e-1 \end{gathered}$$

Thus, ${\int }_1^{\infty }\frac{e^{{\displaystyle \frac{1}{k}}}}{k^2}dk$ converges.

Hence the given series converges.