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I'm trying to find the end behavior of $$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}$$ using the limit comparison test, but I'm having a hardtime finding the comparing equation. I would appreciate if someone could either give me advice to finding the comparing equations and/or the equation of the this problem.

Thanks in advance.

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    $\begingroup$ In the limit comparison test, you usually compare to something simpler. You get something simpler if you drop lower order terms. Try that approach. $\endgroup$
    – Hans Engler
    Apr 4, 2013 at 19:25

4 Answers 4

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$n^2+1 \le 2 n^2$, so $\frac{1}{\sqrt{n^2+1}} \ge \frac{1}{\sqrt{2 n^2}} = \frac{1}{\sqrt{2}} \frac{1}{n}$.

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  • $\begingroup$ I'm not sure how this is applicable could you explain? $\endgroup$
    – David
    Apr 4, 2013 at 19:40
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    $\begingroup$ Just use the comparison test directly. The harmonic series $\frac{1}{n}$ is divergent, and your series is, term by term, greater than the harmonic series, hence it diverges too. $\endgroup$
    – copper.hat
    Apr 4, 2013 at 19:42
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The one that came first to my mind was:

$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}$ $>$ $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+2n+1}}$ $=$

$=$ $\sum_{n=1}^{\infty}\frac{1}{{n+1}}$ $=$ $\infty$ $\hspace{99mm} \blacksquare$

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HINT: $\sqrt{n^2+1}$ is approximately $n$.

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$$\lim_{n\to\infty}\dfrac{\frac1{\sqrt{n^2+1}}}{\frac1n}=1.$$

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