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I have this limit but I cant solve it. I think that it is possible to solve with Stolz, but I can't see how.

$$ \lim_{n\to\infty}\left[\frac{1}{\sqrt{n²+2}}+\frac{1}{\sqrt{n²+4}}+\frac{1}{\sqrt{n²+6}} + \dots + \frac{1}{\sqrt{n²+2n}}\right]$$

Any ideas?

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  • $\begingroup$ Perhaps consider a Riemann sum? $\endgroup$
    – user9464
    Nov 5, 2016 at 15:04
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    $\begingroup$ Either you forgot the term $\frac{1}{\sqrt{n^2+6}}$ or $2n$ should be $2^n$... $\endgroup$ Nov 5, 2016 at 15:08

2 Answers 2

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See that

$${n\over\sqrt{n^2+2n}}\le \frac{1}{\sqrt{n²+2}}+\frac{1}{\sqrt{n²+4}}+\frac{1}{\sqrt{n²+6}} + \dots + \frac{1}{\sqrt{n²+2n}} \le {n\over\sqrt{n^2+2}}$$

And the limits of both ${n\over\sqrt{n^2+2n}}$ and ${n\over\sqrt{n^2+2}}$ is $1$.

Therefore your limit is $1$.

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  • $\begingroup$ and why $\frac{n}{sqrt(n²+2n)} >=sumofall$? $\endgroup$
    – Tlaloc-ES
    Nov 5, 2016 at 15:10
  • $\begingroup$ You can also use $n^2\le n^2+2k\le (n+1)^2$, so your sum is between $\frac{n}{n+1}$ and $\frac{n}{n}$... $\endgroup$ Nov 5, 2016 at 15:13
  • $\begingroup$ @eslop It isn't, it is less than the sum of all because all terms are greater than the first one, and there are $n$ terms $\endgroup$
    – Astyx
    Nov 5, 2016 at 15:16
  • $\begingroup$ @EricTowers Yes but $n\times \text{[smallest term]} \le \text{[sum of all terms]} \le n\times \text{[greatest term]}$ $\endgroup$
    – Astyx
    Nov 5, 2016 at 15:25
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over \root{n^{2} + 2k}} & = \lim_{n \to \infty}\sum_{k = 0}^{n - 1}{1 \over \root{n^{2} + 2k + 2}} = {1 \over \root{2}}\,\lim_{n \to \infty} \sum_{k = 0}^{n - 1}{1 \over \bracks{k + \pars{n^{2}+ 2}/2}^{1/2}} \\[5mm] & = {1 \over \root{2}}\,\lim_{n \to \infty} \bracks{\Phi\pars{1^{-},{1 \over 2},{n^{2} + 2 \over 2}} - \Phi\pars{1^{-},{1 \over 2},{n^{2} + 2 \over 2} + n}} \end{align} where we used a Lerch Zeta Function $\ds{\Phi}$ identity. $\ds{\Phi\pars{1^{-},s,a}}$ satisfies the asymptotic expression: $$ \Phi\pars{1^{-},s,a} \sim {1 \over \pars{s - 1}\, a^{s - 1}}\quad\mbox{as}\quad a \to \infty $$


Then, \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over \root{n^{2} + 2k}} & = {1 \over \root{2}}\,\lim_{n \to \infty}\pars{% -2\root{n^{2} + 2 \over 2} + 2\root{{n^{2} + 2 \over 2} + n}} \\[5mm] & = \lim_{n \to \infty}\pars{\root{n^{2} + 2n + 2} - \root{n^{2} + 2}} \\[5mm] & = \lim_{n \to \infty}{2 \over \root{1 + 2/n + 2/n^{2}} + \root{1 + 2/n^{2}}} = \bbx{\ds{1}} \end{align}

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