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I am trying to determine whether the series$$\sum_{n=1}^\infty{\frac{x}{n(1+nx^2)}}$$is uniformly convergent on the real numbers, and am really stuck. I have tried using the Weierstrass $M$ test but cannot seem to find any series greater than this which is convergent. I have looked at comparing it to series which have $1/n^k$ in particular. Are there any other methods I can use? Or am I just using the Weierstrass M test incorrectly?

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  • $\begingroup$ uniformly convergent where? $\endgroup$ – zhw. Mar 15 '18 at 22:52
  • $\begingroup$ @zhw Sorry, should have clarified. On the reals. $\endgroup$ – Flose Mar 15 '18 at 22:56
  • $\begingroup$ You are not using it incorrectly. In this case a different method is preferable. $\endgroup$ – DanielWainfleet Mar 16 '18 at 0:45
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It is easy to see that$$\max_{x\in\mathbb R}\frac x{n(1+nx^2)}=\frac1{2n^{3/2}}.$$This is so because$$\left(\frac x{n(1+nx^2)}\right)'=0\iff\frac{1-nx^2}{n(1+nx^2)^2}=0\iff x=\pm\frac1{\sqrt n}$$and$$f\left(\pm\frac1{\sqrt n}\right)=\frac1{2n^{3/2}}.$$Now, you can apply the Weierstrass $M$-test with this upper bound.

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  • $\begingroup$ Or simply use the AM-GM inequality as in the solution I posted. ;-)) $\endgroup$ – Mark Viola Mar 15 '18 at 23:04
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From the AM-GM Inequality, we have $1+nx^2\ge 2\sqrt{(1)(nx^2)}=2n^{1/2}|x|$.

Therefore, we can write

$$\begin{align} \left|\sum_{n=1}^\infty \frac{x}{n(1+nx^2)} -\sum_{n=1}^N \frac{x}{n(1+nx^2)}\right|&=\left|\sum_{n=N+1}^\infty \frac{x}{n(1+nx^2)}\right|\\\\ &\le \sum_{n=N+1}^\infty \frac{1}{2n^{3/2}} \end{align}$$

Inasmuch as $\sum_{n=1}^\infty \frac{1}{n^{3/2}}<\infty$, then for any $\epsilon>0$, there exists an number $N$ independent of $x$, such that $\left|\sum_{n=1}^\infty \frac{x}{n(1+nx^2)} -\sum_{n=1}^N \frac{x}{n(1+nx^2)}\right|<\epsilon$, which establishes the uniform convergence.

And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. If it wasn't useful, I am more that happy to delete. Just let me know please. $\endgroup$ – Mark Viola May 8 '18 at 14:59

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