Question

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

Answer 1

Given:

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

So, the value is,

`=(1)/(p^2)\sum_{n\mathop{=}1}^(\infty)(1)/(n^2+3n+1)`

Apply telescoping series

`\sum_{n\mathop{=}1}^(\infty)(1)/(n^2+3n+1)=(1)/(2)`

The value is:

`\begin{gathered} =(1)/(p^2)*(1)/(2) \\ \\ =(1)/(2p^2) \end{gathered}`

It is a series of converges.

`\begin{gathered} =(1)/(2p^2) \\ \\ =(1)/(2(2)^2) \\ \\ =(1)/(2*4) \\ \\ =(1)/(8) \end{gathered}`