Q. 22 - Calculus

Question

Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer. 

             $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{k}{1+k}\right)}^{k^2}$

Step-by-Step Solution

Answer

Ans:  The divergence test fails divergent because,  $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{k}{1+k}\right)}^{k^2}=0$

Step 1. Given information.

given,

       $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{k}{1+k}\right)}^{k^2}$

Step 2. The objective is to use the divergence test to analyze the given series.

The divergence test states that if the sequence $\left\{a_k\right\}$ does not converge to zero, then the series $\sum _{k=1}^{\mathrm{\infty }} a_k$ diverges.

The value of the sequence $\left\{a_k\right\}=\left\{{\left(\frac{k}{1+k}\right)}^{k^2}\right\}$ is:

 

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \left({\left(\frac{k}{1+k}\right)}^{k^2}\right)=0\\ \end{array}$

The divergence test fails divergent because, $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{k}{1+k}\right)}^{k^2}=0$