Q. 60 - Calculus

Question

(a) Show that the series $\sum _{k=1}^{\infty }\frac{\ln k}{k^2}$ converges.

(b) Use the result from part (a) to show that $\sum _{k=1}^{\infty }{\displaystyle \frac{\ln k}{k^n}}$ converges for every integer n > 2.

Step-by-Step Solution

Answer

Part (a) The given series converges.

Part (b) By using the result from part (a) the given series converges for every integer n > 2.  

Part (a) Step 1. Given Information.

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

Part (a) Step 2. Showing that the given series converges.

To show that the given series converges we will use the comparison test.

Let $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\ln k}{k^2}\mathrm{and} \sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^2}.$

So, $0\le \frac{\ln k}{k^2}\le \frac{1}{k^2}.$

Now, $\sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^2} \mathrm{is} \mathrm{of} \mathrm{the} \mathrm{form} \sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^p}.$

If p > 1  then the series converges if p < 1 then the series diverges.

Here $p=2>1$ thus, $\sum _{\mathrm{k}=1}^{\infty }{b}_{\mathrm{k}}=\sum _{k=1}^{\infty }\frac{1}{k^2}$ converges.

By the comparison test, $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\ln k}{k^2}$  also converges.

Thus, the given series converges.

Part (b) Step 1. Showing that the given series converges for every integer n > 2.

As we have shown in part (a) that if p > 1  then the series converges if p < 1 then the series diverges. As it is given that n > 2,  so the given series converges for every integer n > 2.