Q. 41 - Calculus

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. 

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

Step-by-Step Solution

Answer

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

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 $\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$ is convergent. Hence, by integral test, the series $\sum _{k=1}^{\infty }{k}^2{e}^{-k^3}$ is convergent .