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!

  • $\begingroup$ 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. $\endgroup$ – Ali May 28 '13 at 19:37
  • $\begingroup$ @Ali: I agree, except it is really every term after the first. $\endgroup$ – Jonas Meyer May 28 '13 at 19:38
  • $\begingroup$ @JonasMeyer: Oops, I missed that tiny initial term :) $\endgroup$ – Ali May 28 '13 at 19:41
  • $\begingroup$ Try to use this technique. $\endgroup$ – 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 $$ M_n = \frac{1}{2^n-2}.$$

  • $\begingroup$ Did you notice that the domain is $(-2,\infty)$? $\endgroup$ – Jonas Meyer May 28 '13 at 19:31
  • $\begingroup$ Jonas, if the domain is $(-2,\infty)$ then perhaps: $|\frac{(-1)^n}{x+2^n}| \leq |\frac{1}{x+2^n}| \leq |\frac{1}{2^n - 2}|$? $\endgroup$ – TheNotMe May 28 '13 at 19:35
  • $\begingroup$ @TheNotMe: That's right. Of course, the first term is unbounded, but that isn't relevant for convergence. Just throw out the first term to apply the test. (Interesting that now the bound in this hint was first in the OP's comment.) $\endgroup$ – Jonas Meyer May 28 '13 at 19:37
  • $\begingroup$ But does this $M_n$ converge? I wasn't able to prove that it converges to be honest! $\endgroup$ – TheNotMe May 28 '13 at 19:50
  • $\begingroup$ On other thoughts, by the ratio test we get $L=0$ which means it converges. Correct? $\endgroup$ – TheNotMe May 28 '13 at 19:52

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