Q. 44 - 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. 

$\sum _{k=1}^{\infty }\frac{2^k}{k^2}$

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty }\frac{2^k}{k^2}$ is Divergent.

Step 1. Given information

We are given, 

$\sum _{k=1}^{\infty }\frac{2^k}{k^2}$

Step 2. Checking the Convergence and Divergence

The value of the sequence $\left\{a_k\right\}=\left\{\frac{2^k}{k^2}\right\}$ is

$\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{2^k}{k^2}\right)$

Using L'Hopital's Rule,

$=\infty$

The sequence $\left\{a_k\right\}=\left\{\frac{2^k}{k^2}\right\}$ does not converge to zero, therefore the series $\sum _{k=1}^{\infty }\frac{2^k}{k^2}$ diverges.

By divergence test, the series $\sum _{k=1}^{\infty }\frac{2^k}{k^2}$ is divergent.