Answer to Question #263263 in Calculus for Sayem

Question

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

Thank you so much.

Answer 1

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