Find set on which the series $\sum \frac{1}{n(x^2+n)}$ converge uniform. My solution is as follows $|1/(x^2+n)|≤1$ so that $|1/n(x^2+n)|\leq1/n$. Since $\sum1/n$ converges to zero as n goes to infinity , then by Weierstrass test the series converges uniform. Am I in the right track?, I don’t know how I can get values of $x$ for which the given series is uniform convergent. Thanks for any kind of help.
Uniform convergence of $\sum \frac{1}{n(x^2+n)}$
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real-analysis
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3The [harmonic series](http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)) $\sum 1/n$ is the classic example of a series that doesn't converge, despite its terms approaching zero. – 2012-01-07
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0@neemy: I used LaTeX to type the equation. Check if this is what you would like to ask. – 2012-01-07
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1I find it hard to believe that $\sum\frac{x^2+n}{n}$ is intended. – 2012-01-07
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1I agree with André. Did you mean $\sum\frac{1}{n(x^2 + n)}$? About your reasoning: you have $x^2 + n \geq 1$, and so $(x^2 + n)/n \geq 1/n$, which is the opposite of what you have. What does that suggest? – 2012-01-07
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0@Dylan: Maybe I made a mistake when I edited the post. I am sorry. – 2012-01-07
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3@Paul Don't be! I think what you typeset was the only available interpretation of the symbols that were there. – 2012-01-07
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1If it is indeed $\frac{1}{n(x^2+n)}$, applying the $M$-test should work well. – 2012-01-07
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2@neemy: Please try to write formulas correctly and unambiguously. Your second version was a little better, but "$1/a(b)$" does not make it clear whether you want $\frac{1}{ab}$ or $\frac{1}{a}\cdot b$. Please either use LaTeX fractions, e.g. `$\frac{1}{n(x^2+n)}$` to render $\frac{1}{n(x^2+n)}$, or use parentheses correctly, e.g. 1/(n(x^2+n)), until you get the hang of LaTeX. – 2012-01-07
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0Just taking a look at $\zeta(2)$ would also help. – 2012-01-07