series $\frac{3}{1}+\frac{4}{2^p}+\frac{5}{3^p}+\frac{6}{4^p}+\cdot\cdot\cdot$

Question

For what p series:$$\frac{3}{1}+\frac{4}{2^p}+\frac{5}{3^p}+\frac{6}{4^p}+\cdot\cdot\cdot$$ is convergent?

Answer:

$\frac{3}{1}+\frac{4}{2^p}+\frac{5}{3^p}+\frac{6}{4^p}+\cdot\cdot\cdot=\frac{1+2}{1}+\frac{2+2}{2^p}+\frac{3+2}{3^p}+\frac{4+2}{4^p}+\cdot\cdot\cdot=\sum\limits_{i=1}^\infty \frac{n+2}{n^p}$

now what should I do?

Answer 1

$$\begin{align*}\sum_{n=1}^\infty \dfrac{n+2}{n^p} & = \sum_{n=1}^\infty \dfrac{1}{n^{p-1}} + \sum_{n=1}^\infty \dfrac{2}{n^p} \\ & = \sum_{n=1}^\infty n^{1-p} + 2\sum_{n=1}^\infty n^{-p}\end{align*}$$

This is the sum of two p-series. You need $p>2$ for it to converge.

Answer 2

Almost done:

$\dfrac{n}{n^p} +2\dfrac{1}{n^p}.$

$a_n: = \dfrac{n}{n^p}= \dfrac{1}{n^{p-1}}$.

$\sum a_n$ converges for $p-1>1$, or $p>2.$

$b_n= 2\dfrac{1}{n^p}$ ;

$\sum b_n$ converges for $p>1.$

Hence ?