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Consider the series $\sum_{k=1}^\infty \frac{1}{x^2+k^2}$

Show that this series converges to a continuous function that is defined for all $x ∈ R$.

I'm unsure how to approach this. I'm wondering if I can use the M-test to show that this series converges. Since the power series is continuous, it would have to converge to a continuous function, $f$, if it does converge. Is this correct?

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  • $\begingroup$ $ \frac{1}{x^2+k^2} \le \frac{1}{k^2}$. $\endgroup$
    – lhf
    Dec 8, 2015 at 0:17
  • $\begingroup$ Could you elaborate a little more on this? I'm unsure what to do with this? $\endgroup$
    – user282934
    Dec 8, 2015 at 0:18
  • $\begingroup$ $\displaystyle\sum_{k=-\infty}^\infty\frac1{x^2+k^2}~=~\dfrac\pi x~\coth(\pi x)$ $\endgroup$
    – Lucian
    Dec 8, 2015 at 0:19

1 Answer 1

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You're right about using the Weierstrass M-test:

  • $\dfrac{1}{x^2+k^2}$ is uniformly bounded on $x$ because $ \dfrac{1}{x^2+k^2} \le \dfrac{1}{k^2}$ .

  • $\displaystyle \sum_k \dfrac{1}{k^2} < \infty$ (you may even know that the value is $\dfrac{\pi^2}{6}$).

  • Apply the Weierstrass M-test to conclude that the original series converges uniformly.

  • Use that if a series of continuous functions converges uniformly, then the limit is a continuous function.

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