Find all values of x for which the series converges.For these values of x, write the sum of the series as a function of x.
E)The series diverges for all x.

Question

Find all values of x for which the series converges.For these values of x, write the sum of the series as a function of x. 
E)The series diverges for all x.

Quizplus · Accepted Answer

The series $\sum_{n=0}^{\infty} \frac{x^n}{n^2+1}$ converges for all values of x because the terms of the series are positive and decreasing. To see this, note that for $x \geq 0$, the ratio of successive terms is $$\frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(n+1)^2+1} \cdot \frac{n^2+1}{x^n} = \frac{x}{n^2+2n+2+\frac{1}{x^n}} \leq x,$$ and for $-1<x<0$, we have $$\frac{a_{n+1}}{a_n} = \frac{|x|^{n+1}}{(n+1)^2+1} \cdot \frac{n^2+1}{|x^n|} \leq |x|.$$ Therefore, the series converges by the ratio test for all $x \in [-1,1]$.

To find the sum of the series, let $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n^2+1}$. The function $f(x)$ is continuous on $[-1,1]$ since the series converges uniformly on any compact subset of $[-1,1]$. Furthermore, it is differentiable on $(-1,1)$ since the series converges uniformly on any closed subinterval of $(-1,1)$.

Differentiating term by term, we have $$f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{x^n}{n^2+1}ight) = \sum_{n=0}^{\infty} \frac{nx^{n-1}}{n^2+1} = \sum_{n=0}^{\infty} \frac{nx^{n-1}}{(n+1/2)^2+1/4}.$$ Multiplying the numerator and denominator by $2(n+1/2)$, we have $$\frac{nx^{n-1}}{(n+1/2)^2+1/4} = \frac{2(n+1/2)x^{n}}{2(n+1/2)^2+1} - \frac{(n+1/2)x^{n-1}}{2(n+1/2)^2+1},$$ so $$f'(x) = 2x \sum_{n=0}^{\infty} \frac{(n+1/2)x^{n-1}}{2(n+1/2)^2+1} - \sum_{n=0}^{\infty} \frac{(n+1/2)x^{n-1}}{2(n+1/2)^2+1}.$$ The second series can be evaluated using the substitution $m = n+1/2$, giving $$\sum_{n=0}^{\infty} \frac{(n+1/2)x^{n-1}}{2(n+1/2)^2+1} = \sum_{m=1}^{\infty} \frac{mx^{m-1/2}}{2m^2+1}.$$ The series on the right-hand side is the derivative of the series $\sum_{m=1}^{\infty} \frac{x^{m-1/2}}{2m^2+1}$, which can be obtained by multiplying the series $\sum_{m=1}^{\infty} \frac{x^{m-1/2}}{m}$ by the power series for $\sqrt{x}$, namely $\sum_{k=0}^{\infty} \binom{1/2}{k} x^{k}$. Thus, we have $$\sum_{m=1}^{\infty} \frac{mx^{m-1/2}}{2m^2+1} = \frac{d}{dx}\left(\sum_{m=1}^{\infty} \frac{x^{m-1/2}}{2m^2+1}ight) = \frac{d}{dx}\left(\frac{1}{x^{1/2}}\sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m}}{2m^2+1}ight)$$ $$= -\frac{1}{2x^{3/2}}\sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m}}{2m^2+1} + \frac{1}{x^{1/2}}\sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m-2}}{2m^2+1}.$$ Therefore, we have $$f'(x) = \frac{1}{x^{1/2}}\left(2\sum_{n=0}^{\infty} \frac{(n+1/2)x^{n}}{2(n+1/2)^2+1} - \frac{1}{x}\sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m}}{2m^2+1} + \sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m-2}}{2m^2+1}ight)$$ $$= \frac{1}{x^{1/2}}\left( \frac{1}{1+x}\sum_{n=0}^{\infty} \frac{x^{n}}{n^2+1} + \frac{1}{2}\sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m}}{m^2+\frac{1}{2}} + \sum_{m=1}^{\infty} \frac{(x^{1/2})^{2m-2}}{2m^2+1}ight).$$ Since $\sum_{n=0}^{\infty} \frac{x^{n}}{n^2+1}$ converges uniformly on any compact subset of $[-1,1]$ and $f'(x)$ is continuous on $[-1,1]$, we can integrate term by term to obtain $$f(x) = f(0) + \int_{0}^{x} f'(t) \, dt = \int_{0}^{x} f'(t) \, dt = \int_{0}^{x} \frac{1}{t^{1/2}} G(t) \, dt,$$ where $G(t)$ is the right-hand side of the expression for $f'(x)$. This integral can be evaluated using integration by parts and the fact that $\lim_{t 	o 0^+} t^{1/2} G(t)$ exists, giving $$f(x) = \frac{\pi}{2} - \frac{\pi}{2}\frac{\exp(-x^{1/2})}{1-\exp(-\pi)} -\frac{1}{2}\int_{0}^{x^{1/2}} \frac{\exp(t)}{t} \sum_{m=1}^{\infty} \frac{(xt^2)^{m-1/2}}{m^2+\frac{1}{2}} \, dt $$ $$+ \int_{0}^{x^{1/2}} \frac{\sin(t)\,dt}{t^{3/2}} \left(\sum_{m=1}^{\infty} \frac{(xt^2)^{m}}{m^2+\frac{1}{2}} - \frac{xt^2}{2}ight).$$

Find All Values of X for Which the Series Converges