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Does $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}$ converge (absolutely)?

I've already tried the Ratio test and the Limit comparison test, but the series failed both...

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  • $\begingroup$ Have you (carefully) tried a comparison with the series $\sum 1/n$? $\endgroup$ – Old John Dec 6 '13 at 22:19
  • $\begingroup$ I prefer Comparison, but Limit Comparison (with $\sum \frac{1}{n}$) works OK. $\endgroup$ – André Nicolas Dec 6 '13 at 22:22
  • $\begingroup$ Since you have the word "absolute" in the question, are you sure that the series does not have a factor $(-1)^n$ in it? $\endgroup$ – imranfat Dec 6 '13 at 23:57
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Note that $$\dfrac1{\sqrt{n^2+1}} \geq \dfrac1{n+1}$$ and recall that $$\sum_{n=0}^{\infty} \dfrac1{n+1}$$ diverges.

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Hint:

$$\frac1{\sqrt{n^2+1}}\ge\frac1{\sqrt2\,n}\;\ldots$$

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