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how to check if this converge? $$\sum_{n=1}^\infty a_n$$

$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$

what i did is to show that:

$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$

and we know that: $$\sum_{n=1}^\infty b_n$$ doesnt converge cause $$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$ doesnt converge

so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge

but i know the final answer is that it does converge so what am i doing wrong?

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marked as duplicate by AlexR, Shaun, Hippalectryon, Ali Caglayan, Claude Leibovici Jan 22 '15 at 11:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You'll want to fix your notation if you want to work with this. $$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} - \sqrt{i+1}$$ is presumably what you really want to study. $\endgroup$ – AlexR Jan 22 '15 at 9:35
  • $\begingroup$ I take it you mean to ask if the sequence:$$S_n=\sum_{k=1}^n \left(2\sqrt{k}-\sqrt{k-1}-\sqrt{k+1}\right)$$ converges as $n\to \infty $? $\endgroup$ – Conrad Turner Jan 22 '15 at 9:39
  • $\begingroup$ its not exactly a dpulicate cuase I'm not looking for the answer, I want to know what is wrong with my way of solving $\endgroup$ – Daniel Katzan Jan 22 '15 at 9:47
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Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so $$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1}$$


NB that your estimate contains an error in the last step, [fixed now] $$2 \sqrt n - 2 \sqrt{n+1} = 2 ( \sqrt n - \sqrt{n+1}) \ne 2 (\sqrt n + \sqrt {n+1})$$ Your estimate only proves $S_n > 0$, wich doesn't show anything on its own.

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  • $\begingroup$ I though of this way, but I'm looking for the reason that my approach is incorrect? $\endgroup$ – Daniel Katzan Jan 22 '15 at 9:45
  • $\begingroup$ @DanielKatzan $2 \sqrt n - 2\sqrt{n+1} \ne -2 (\sqrt n + \sqrt{n+1})$ in your last step there's an error. $\endgroup$ – AlexR Jan 22 '15 at 9:46
  • $\begingroup$ i fixed that mistake $\endgroup$ – Daniel Katzan Jan 22 '15 at 9:49
  • $\begingroup$ @DanielKatzan Not exactly. You still miss the brackets and $\lim_{n\to\infty} \sqrt n - \sqrt{n+1} = 0$ so you don't show anything as I said. $\endgroup$ – AlexR Jan 22 '15 at 9:50
  • $\begingroup$ I edited again. $\endgroup$ – Daniel Katzan Jan 22 '15 at 9:53
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We have

$$2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}=\sqrt n\left(2-\sqrt{1-\frac1n}-\sqrt{1+\frac1n}\right)\\\sim_\infty \sqrt n\left(2-1+\frac1{2n}+\frac1{4n^2}-1-\frac1{2n}+\frac1{4n^2}\right)=\frac1{2n^{3/2}}$$ so the given series is convergent by comparison with a convergent Riemann series.

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  • $\begingroup$ A little overkill since you can compute $S_n$ explicitly ;) $\endgroup$ – AlexR Jan 22 '15 at 9:44
  • $\begingroup$ You're right and I do it on purpose to give different methods;) $\endgroup$ – user63181 Jan 22 '15 at 9:47
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$$ \begin{align} &2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\\[16pt] &=(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})\\[9pt] &=\frac1{\sqrt{n}+\sqrt{n-1}}-\frac1{\sqrt{n+1}+\sqrt{n}}\\ &=\frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}\\ &=\frac2{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1})}\\ &\le\frac1{4(n-1)^{3/2}} \end{align} $$ Thus, $$ \begin{align} \sum_{n=1}^\infty\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right) &\le(2-\sqrt2)+\sum_{n=2}^\infty\frac1{4(n-1)^{3/2}} \end{align} $$ and the series converges by the $p$-test.

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  • $\begingroup$ thx, but i wasnt looking for the answer, I'm looking for the reason my answer is wrong, i tried to use the same method like you, only i got a result where i got something that is bigger then something that doesnt converge by the p-test $\endgroup$ – Daniel Katzan Jan 22 '15 at 15:32
  • $\begingroup$ @DanielKatzan: Well, the title and the first sentence ask to show if the series converges. So, I answered that. I figured by comparing what I did to what you did, you might find where things went wrong. For the most part, everything you said was correct, except your last inequality is slightly wrong: $$-\frac1{\sqrt{n}}\lt-\frac2{\sqrt{n+1}+\sqrt{n}}\lt-\frac1{\sqrt{n+1}}$$ so, with this correction, you have shown that $$2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\gt-\frac1{\sqrt{n}}$$ which is true, but neither proves nor disproves convergence. $\endgroup$ – robjohn Jan 22 '15 at 16:14
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$$\begin{eqnarray*}\sum_{n=1}^{N}\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right)&=&\color{red}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n-1}\right)}+\color{blue}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n+1}\right)}\\&=&\color{red}{\sqrt{N}}+\color{blue}{1-\sqrt{N+1}}\\&=&1-\frac{1}{\sqrt{N}+\sqrt{N+1}},\tag{1}\end{eqnarray*}$$ so, by letting $N\to +\infty$, we get that the original series converges towards $\color{green}{1}$.

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  • $\begingroup$ (+1) The coloring is a bit confusing actually. It would be easier to see if you wrote the blue sum as $$-\sum_{n=2}^{N+1}(\sqrt{n}-\sqrt{n-1})$$ $\endgroup$ – robjohn Jan 22 '15 at 11:16

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