Answer in Calculus for Sayem #263263

1. Find a power series representation centered at x = 0 of the function x² / (4-x²) and determine its interval of convergence.
2. Determine all numbers x so that the series “displaystylesum_{k=0}^u221e” (ln x)^k converges
3. For which p is the series “displaystylesum_{n=0}^u221e” n/ (n²+1)^p convergent?

Thank you so much.

1. We are given f(x)=“frac{x^2}{4-x^2}”

We know “displaystylesum_{n=0}^{infty}x^n=frac{1}{1-x} ,|x|<1” this is equation (i), Where 1 is the first term and x is the common ratio.

Here f(x)=“frac{x^2}{4-x^2}=frac{frac{x^2}{4}}{1-frac{x^2}{4}}” Here first term is x² and common ratio is (x²-3)

Relating this with equation (i)

“frac{x^2}{4-x^2}=frac{x^2}{4}displaystylesum_{n=0}^{infty}(frac{x^2}{4})^n, and |frac{x^2}{4}|<1”

“=displaystylesum_{n=0}^{infty}frac{x^{2n}}{4^{2n}}frac{x^2}{4}”

“=displaystylesum_{n=0}^{infty}frac{x^{2(n+1)}}{4^{(2n+1)}}”

And for interval of convergence,

“|frac{x^2}{4}|<1implies |x^2|<4”

“implies x^2<4”

“-2<x<2”

“therefore” the interval of convergence is (-2,2) and the series is “displaystylesum_{n=0}^{infty}frac{x^{2(n+1)}}{4^{(2n+1)}}”

2. The given series is “displaystylesum_{k=0}^{infty}(ln x)^k”

By D’Alembert’s ratio test

“limlimits_{krarrinfty}frac{a_{k+1}}{a_k}=limlimits_{krarrinfty}frac{(ln x)^{k+1}}{(ln x)^k}”

“=limlimits_{krarrinfty}(ln x)^{k+1-k}”

“=limlimits_{krarrinfty}(ln x)”

“=ln x”

Since it converges, “therefore ln x<1”

“implies 0<x<e”

The values of x that satisfies the series is “0<x<e”

3. “displaystylesum_{n=0}^{infty}frac{n}{(n^2+1)^p}” “implies” Consider the series “displaystylesum_{n=0}^{infty}frac{1}{(n^{2p-1})}”

Based on p series test

If 2p-1>1 then the series converges

2p>1+1=2

p>1

The series is convergent for p>1