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How do I find using the Squeeze theorem

$$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;,$$

using the fact that

$$ \lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}}=1.$$

Thank you very much for your help,

C.G

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Note that $$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\leq \frac{n}{\sqrt{n^2+1}}$$ because $\sqrt{n^2+1}\leq\sqrt{n^2+k}$ for $k\geq1$.

On the other hand, by a similar reasoning, using that $\sqrt{n^2+n}\geq\sqrt{n^2+k}$ for $k\geq1$: $$\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}$$

so in the end, $$\frac{n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}}\leq \frac{n}{\sqrt{n^2+1}}$$ and use that $$\lim_{n} \frac{n}{\sqrt{n^2+n}}=1=\lim_{n} \frac{n}{\sqrt{n^2+1}}$$

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  • 1
    $\begingroup$ Slight mistake in the numerator of the last summation. $\endgroup$ – adfriedman Feb 14 '17 at 17:50

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