Answer to Question #263263 in Calculus for Sayem

Question #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 "\\displaystyle\\sum_{k=0}^\u221e" (ln x)k converges
  3. For which p is the series "\\displaystyle\\sum_{n=0}^\u221e" n/ (n2+1)p convergent?

Thank you so much.


1
Expert's answer
2021-11-10T11:49:34-0500

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

We know "\\displaystyle\\sum_{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}\\displaystyle\\sum_{n=0}^{\\infty}(\\frac{x^2}{4})^n, and \\ |\\frac{x^2}{4}|<1"

"=\\displaystyle\\sum_{n=0}^{\\infty}\\frac{x^{2n}}{4^{2n}}\\frac{x^2}{4}"

"=\\displaystyle\\sum_{n=0}^{\\infty}\\frac{x^{2(n+1)}}{4^{(2n+1)}}"

And for interval of convergence,

"|\\frac{x^2}{4}|<1\\implies |x^2|<4"

"\\implies x^2<4"

"-2<x<2"

"\\therefore" the interval of convergence is (-2,2) and the series is "\\displaystyle\\sum_{n=0}^{\\infty}\\frac{x^{2(n+1)}}{4^{(2n+1)}}"

2. The given series is "\\displaystyle\\sum_{k=0}^{\\infty}(ln\\ x)^k"

By D'Alembert's ratio test

"\\lim\\limits_{k\\rarr\\infty}\\frac{a_{k+1}}{a_k}=\\lim\\limits_{k\\rarr\\infty}\\frac{(ln\\ x)^{k+1}}{(ln\\ x)^k}"

"=\\lim\\limits_{k\\rarr\\infty}(ln\\ x)^{k+1-k}"

"=\\lim\\limits_{k\\rarr\\infty}(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. "\\displaystyle\\sum_{n=0}^{\\infty}\\frac{n}{(n^2+1)^p}" "\\implies" Consider the series "\\displaystyle\\sum_{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







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