Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm attempting to learn some complex analysis on my own, and I've run across a curious question.

My book says to "discuss" the uniform convergence of the series $\displaystyle\sum_{k=1}^\infty\frac{x}{k(1+kx^2)}$ for real values $x$.

I interpret this to determine the values of $x$ where the series is uniformly convergent, and I assume that means when the sequence of partial sums is uniformly convergent.

I define a sequence of functions $\{s_n(x)\}$ defined by $$ s_n(x)=\sum_{k=1}^n\frac{x}{k(1+kx^2)}. $$ Now for any $\epsilon>0$, I think I would like to find an $n_0$ such that for all $m\geq n\geq n_0$, $$ |s_m(x)-s_n(x)|=\left|\sum_{k=n+1}^m\frac{x}{k(1+kx^2)}\right|<\epsilon. $$

Am I correct in thinking that the $x$ which satisfy this for all $\epsilon$ will be the $x$ where the series is uniformly continuous? If so, how could I do this? I hope I have not interpreted the problem wrongly. Thanks kindly for your aid.

share|improve this question
add comment

2 Answers 2

By the arithmetic-geometric mean inequality, $(1 + k x^2)/2 \ge |x| \sqrt{k}$, so $\left| \frac{x}{k(1+kx^2)}\right| \le \frac{2}{k^{3/2}}$.

This makes it easier, since $\sum k^{-3/2}$ converges.

share|improve this answer
Thanks marty. Does this imply that the series converges uniformly on all of $\mathbb{C}$? My thinking is since it's absolutely convergent, then it's ordinarily convergent, so I would always be able to find partial sums large enough that their difference is smaller than some given $\epsilon>0$. –  Dedede Jan 30 '12 at 5:31
That inequality fails when $\: k=1 \:$ and $\;\; x \: = \: \frac12 \cdot i \;\;$. $\;\;\;\;\;$ –  Ricky Demer Jan 30 '12 at 8:59
add comment

That means determine the subsets of the complex plane on
which the sequence of partial sums is uniformly convergent.
For example, one conceivable answer would be "exactly the bounded subsets".
"uniform convergence at $x$" does not make sense, and
uniform convergence on $\{x\}$ is equivalent to convergence at $x$.

Perhaps, but the general rule is that if a sequence of continuous functions converges uniformly
on an open set, then the function defined by the limit is continuous on that open set.

share|improve this answer
Thanks. So is there some systematic way to determine on which subsets the sequence of partial sums is uniformly convergent? –  Dedede Jan 30 '12 at 5:13
Basically, find out an "essentially tight" modulus of convergence, where "essentially tight" means whatever you need it to mean to prove that uniform convergence fails on all other sets. $\;$ –  Ricky Demer Jan 30 '12 at 9:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.