# Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}}$ in $(-2,\infty)$ uniformly convergent?

Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}}$ in $(-2,\infty)$ uniformly convergent?

I started by checking if it is pointwise convergent, because if it wasn't then especially it is not uniformly conveergent. But by Leibnitz, it is convergent.

But I got stuck on proving that it is uniformly convergent. Any help would be appreciated!

• I think the fact that the absolute value of each term is smaller than $\frac{1}{2^{n-1}}$ should lead to what you are looking for. – Ali May 28 '13 at 19:37
• @Ali: I agree, except it is really every term after the first. – Jonas Meyer May 28 '13 at 19:38
• @JonasMeyer: Oops, I missed that tiny initial term :) – Ali May 28 '13 at 19:41
• Try to use this technique. – Mhenni Benghorbal May 28 '13 at 21:52

Also a hint: Set $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{x+2^{n}}$. Now consider sequences $x_{k}=-2+\frac{1}{2^{k}}$ and $y_{k}=-2+\frac{1}{2^{k+1}}$. Obviously $|x_{k}-y_{k}|\to 0$ when $k\to\infty$. Now see what happens with $|f(x_{k})-f(y_{k})|=|x_{k}-y_{k}||\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(x_{k}+2^{n})(y_{k}+2^{n})}|$. More precisely, consider the first term in summation...
Hint: Use Weierstrass M-test with $$M_n = \frac{1}{2^n-2}.$$
• Did you notice that the domain is $(-2,\infty)$? – Jonas Meyer May 28 '13 at 19:31
• Jonas, if the domain is $(-2,\infty)$ then perhaps: $|\frac{(-1)^n}{x+2^n}| \leq |\frac{1}{x+2^n}| \leq |\frac{1}{2^n - 2}|$? – TheNotMe May 28 '13 at 19:35
• But does this $M_n$ converge? I wasn't able to prove that it converges to be honest! – TheNotMe May 28 '13 at 19:50
• On other thoughts, by the ratio test we get $L=0$ which means it converges. Correct? – TheNotMe May 28 '13 at 19:52