Question

For the series that follow, determine whether the series converges or diverges. Explain the criteria you are using and why your conclusion is valid.

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

Short answer

The series converges.

Using the root test we can say that the series has positive terms. Then apply limits and find the solution.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

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

Step 2. Use the root test.

According to the Root test,

$\sum _{k=1}^{\infty }{a}_k$is the series, if $L=\underset{k\to \mathrm{\infty }}{lim} {\left(a_k\right)}^{1/k}$then,

First is if role="math" localid="1650691303843" $L<1$series converges.

Second is if $L>1$series diverges.

Third is if $L=1$the test is inconclusive.

The general term of the series $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{1}{k^2}\right)}^{\sqrt{k}}$is $a_k={\left(\frac{1}{k^2}\right)}^{\sqrt{k}}$.

Since, the series has positive terms, hence it meets the hypothesis of the Root test.

Step 3. Use the limit.

The value of $\underset{k\to \mathrm{\infty }}{lim} {\left(a_k\right)}^{1/k}$is given below,

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

As the value of the limit is $1$.

Therefore, the series $\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{1}{k^2}\right)}^{\sqrt{k}}$converges.