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Show that the series $$\sum_{n=0}^\infty \frac{1}{1+n^2x}$$ converges uniformly on $[a,\infty)$ for any $a>0$ but does not converge uniformly on $(0,\infty)$.

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closed as off-topic by Jonas Meyer, Tunk-Fey, Thursday, Jyrki Lahtonen, J. W. Perry Aug 18 at 5:58

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2 Answers 2

Hints:

Set $f_n(x)={1\over 1+n^2 x}$.

For uniform convergence on $[a,\infty)$ with $a>0$:

Note $|f_n(x)|\le {1\over 1+a n^2}$. Consider using the Weierstrass M-test.


On the interval $(0,\infty)$:

For $n>0$, let $x_n=1/n^2$. Note $f_n(x_n)=1/2$ for each $n$. Could the sequence $(f_n)$ converge uniformly to zero on $(0,\infty)$? What does this tell you about uniform convergence of the series $\sum\limits_{n=0}^\infty f_n$ on $(0,\infty)$?

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I forgot, but isn't there a theorem saying that if the series converges uniformly in $(a,b)$ then it converges at the end points as well? In this case it is simple to just show that $f_n(0)=1$ for all $n$ and therefore there is no convergence at $0$ so there cannot be a uniform convergence at $(0,\infty)$. –  Asaf Karagila May 16 '12 at 23:27

Use the Weierstrass M test: $1/(1 + n^2 x) \leq 1/n^2$ for all $x > 0$. Let $M_n = 1/n^2$. We know $\sum 1/n^2$ is convergent by a p-series test, hence by the M-test the original series is uniformly convergent. Note that if $x = 0$, then the sum is just $\sum 1$ which is clearly divergent. Source(s): http://en.wikipedia.org/wiki/Weierstrass

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