Question

Use any convergence test from this section or the previous section to determine whether the series in Exercises 31–48 converge or diverge. Explain how the series meets the hypotheses of the test you select.

$\sum _{k=1}^{\infty }\frac{1}{\sqrt{k+1}+\sqrt{k}}$

Short answer

The series $\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$is Convergent.

Step-by-step solution

Step 1. Given information  

We are given,

$\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$

Step 2. Checking the Convergence and Divergence 

Consider the integral ${\int }_{x=1}^{\infty }f\left(x\right)dx={\int }_{x=1}^{\infty }{x}^2{e}^{-x^3}dx$.

Evaluating the integral, integrate by part,

$$\begin{gathered} {\int }_{x=1}^{\infty }f\left(x\right)dx=\underset{k\to \infty }{\lim }{\int }_{x=1}^k{x}^2{e}^{-x^3}dx \\ =\frac{1}{3}\underset{k\to \infty }{\lim }{\int }_{u=1}^{k^3}{e}^{-u}du\text{(Put}{x}^3=u\Rightarrow 3x^{2}dx=du\text{)} \\ =\frac{1}{3}\underset{k\to \infty }{\lim }{\left[-e^{-u}\right]}_1^{k^3} \\ =\frac{1}{3}\underset{k\to \infty }{\lim }\left[-e^{-k^3}+e\right] \\ =\frac{e}{3} \end{gathered}$$

Step 3. Checking the Convergence and Divergence 

Thus, the value of the integral is ${\int }_{x=1}^{\infty }{x}^2{e}^{-x^3}dx=\frac{e}{3}$.

The integral converges. Therefore, the series role="math" localid="1649316120490" $\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$is convergent. Hence, by integral test, the series $\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$is convergent .