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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.

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The harmonic series $\sum 1/n$ is the classic example of a series that doesn't converge, despite its terms approaching zero. – Dylan Moreland Jan 7 '12 at 4:56
I find it hard to believe that $\sum\frac{x^2+n}{n}$ is intended. – André Nicolas Jan 7 '12 at 4:59
I 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? – Dylan Moreland Jan 7 '12 at 5:04
@Paul Don't be! I think what you typeset was the only available interpretation of the symbols that were there. – Dylan Moreland Jan 7 '12 at 5:08
@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. – Jonas Meyer Jan 7 '12 at 5:50

Hint: The estimate ${1\over x^2+n}\le 1$ is "too much"; you are throwing away a term that actually helps you (the $n$). Estimate with ${1\over x^2+n}\le {1\over n}$ instead.

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