Uniform Convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$

Question

Uniform convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$

Using Dirichlet Test, it can be shown that the it uniformly converges for $\mathbb{R}$ \ $\{0\}$.

In $x=0$ there is obviously a pointwise convergence to 0.

However, I'm struggling to make a conclusion for $\mathbb{R}$.

If you could help me find a conclusion or provide another direction, it would be appreciated.

Answer 1

In short: here, you shouldn't have to worry about $0$ in the first place.

When you apply the Dirichlet test, take $$ a_n(x) = (-1)^{n-1}, \qquad b_n(x) = \frac{x^2}{(1+x^2)^n} $$ for $x\in\mathbb{R}$ and $n\geq 1$. Then

• For $M\stackrel{\rm def}{=} 1$, we have $$ \left\lvert \sum_{n=1}^N a_n(x) \right\rvert \leq M $$ for all $N\geq 1$ and $x\in\mathbb{R}$.

• For all $x\in\mathbb{R}$ and $n\geq 1$, $$ b_n(x) \geq b_{n+1}(x) $$

• $\lim_{n\to\infty }\lVert b_n\rVert_\infty = 0$ (uniform convergence of $b_n$ to $0$)

Therefore, the (Uniform) Dirichlet test guarantees that the series $$ \sum_{n=1}^\infty a_n(x)b_n(x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^2}{(1+x^2)^n} $$ converges uniformly on $\mathbb{R}$.

Answer 2

Note \begin{eqnarray} S_N(x)&=&\sum_{n=1}^{N}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}=\frac{x^2}{1+x^2}=\sum_{n=1}^{N}(-1)^{n-1}\frac{1}{(1+x^2)^{n-1}}\\ &=&\frac{x^2}{1+x^2}\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{1+\frac{1}{1+x^2}}=x^2\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{x^2+2}\\ \end{eqnarray} and hence \begin{eqnarray} \left|S_N(x)-\frac{x^2}{x^2+2}\right|&=&\frac{x^2}{(1+x^2)^{N}(x^2+2)}. \end{eqnarray} Let $$ f_N(x)=\frac{x}{(1+x)^{N}(x+2)}, x\ge0. $$ Solveing $f'(x)=0$ gives $x=x_N\equiv\frac{\sqrt{N^2+1}-N+1}{N}$ and noting $f_N''(x_N)<0$. Therefore $f_N(x)$ attains the max at $x=x_N$. Thus \begin{eqnarray} &&\max\left|S_N(x)-\frac{x^2}{x^2+2}\right|=\max f_N(x^2)=\max_{x\ge0}f_N(x)\\ &=&f_N(x_N)=\frac{\sqrt{N^2+1}-N+1}{\sqrt{N^2+1}+N+1}\left(\frac{N}{\sqrt{N^2+1}+1}\right)^N\\ &\le&\frac{\sqrt{N^2+1}-N+1}{\sqrt{N^2+1}+N+1}\\ &\le&\frac{2}{\sqrt{N^2+1}+N+1}\\ &\to&0 \end{eqnarray} as $N\to\infty$ and thence $\{S_N(x)\}$ converges uniformly to $\frac{x^2}{x^2+2}$.