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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.

2 Answers 2

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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.

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    That inequality fails when $\: k=1 \:$ and $\;\; x \: = \: \frac12 \cdot i \;\;$. $\;\;\;\;\;$2012-01-30
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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.

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    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. $\;$2012-01-30