Question

Please solve this question for me Note: the value of p=2

Answer 1

Given:

$\sum_{n\mathop{=}1}^{\infty}\frac{(np)^{-2}}{\frac{2}{n^2}+\frac{3}{n}+1}$

Find-:

Find the formula for n partial sum, and determine whether the series converges or diverges.

Explanation-:

The value is:

$\begin{gathered} =\sum_{n\mathop{=}1}^{\infty}\frac{(np)^{-2}}{\frac{2}{n^2}+\frac{3}{n}+1} \\ \\ =\sum_{n\mathop{=}1}^{\infty}\frac{1}{n^2p^2(\frac{1}{n^2}+\frac{3}{n}+1)} \\ \\ =\sum_{n\mathop{=}1}^{\infty}\frac{1}{p^2(\frac{n^2}{n^2}+\frac{3n^2}{n}+n^2)} \\ \\ =\sum_{n\mathop{=}1}^{\infty}\frac{1}{p^2(1+3n+n^2)} \end{gathered}$

So, the value is,

$=\frac{1}{p^2}\sum_{n\mathop{=}1}^{\infty}\frac{1}{n^2+3n+1}$

Apply telescoping series

$\sum_{n\mathop{=}1}^{\infty}\frac{1}{n^2+3n+1}=\frac{1}{2}$

The value is:

$\begin{gathered} =\frac{1}{p^2}\times\frac{1}{2} \\ \\ =\frac{1}{2p^2} \end{gathered}$

It is a series of converges.

$\begin{gathered} =\frac{1}{2p^2} \\ \\ =\frac{1}{2(2)^2} \\ \\ =\frac{1}{2\times4} \\ \\ =\frac{1}{8} \end{gathered}$