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I'm having trouble determining the convergence of the series: $$ \sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} $$

I am thinking it doesn't converge and since neither the root test or $$|\frac{a_{k+1}}{a_{k}}|$$ seemed to work for me I would have to use a comparing test

Keep in mind I am not allowed to actually calculate what it converges to.

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    $\begingroup$ It telescopes${}$. $\endgroup$ Dec 7, 2017 at 19:55
  • $\begingroup$ it should give me something like $\sqrt{1} - \sqrt{2} + \sqrt{n +2} - \sqrt{n+1}$ but how does that help me determine the convergence? $\endgroup$
    – John.Doh
    Dec 7, 2017 at 20:06
  • $\begingroup$ @John.Doh: $$\lim_{n\to +\infty}\left[\sqrt{1}-\sqrt{2}+\sqrt{n+2}-\sqrt{n+1}\right]$$ is finite. $\endgroup$ Dec 7, 2017 at 20:18
  • $\begingroup$ The problem is yes while it seems like a duplicate. I can not use the telescope method because I have to determine whether or not it converges by not actually calculating to what it converges(if it converges). Hence the solutions given there are invalid for me. $\endgroup$
    – John.Doh
    Dec 7, 2017 at 20:29

3 Answers 3

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Here's an outline of a solution:

Notice that

$$\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2} = \sqrt{k + 2} - \sqrt{k + 1} - \big(\sqrt{k + 1} - \sqrt{k}\big)$$

is the difference between the right- and left- hand estimates for the derivative of $\sqrt{x}$ at $k + 1$. This has a lot in common with the second derivative of $\sqrt{x}$, which is of the order $x^{-3/2}$. Hence, you may find it very useful to compare your series with

$$\sum_{k = 1}^{\infty} \frac{1}{k^{3/2}}$$

which is easily seen to converge.


Alternatively, as pointed out in comments, the series telescopes.

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  • $\begingroup$ but isn't it $$\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2} = \sqrt{k + 2} - \sqrt{k + 1} - \big(-\sqrt{k} + \sqrt{k + 1}\big)$$ ? $\endgroup$
    – John.Doh
    Dec 7, 2017 at 20:08
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    $\begingroup$ Yes, a typo. The point is the same, though. $\endgroup$
    – user296602
    Dec 7, 2017 at 20:09
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    $\begingroup$ @user296602: but the derivative of root $x$ is $\frac{1}{2x^{1/2}}$. $\endgroup$
    – daulomb
    Dec 7, 2017 at 20:17
  • $\begingroup$ @daulomb Yes, but this is a second difference, which is related to the second derivative. $\endgroup$
    – user296602
    Dec 7, 2017 at 20:35
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For real $0 < x < 1,$ you can check, by squaring, that $$ 1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} $$ You can get the same information from the first few terms of the Taylor series, with explicit remainder.

Using $x = 1/k$ and $x = 2/k$ gives $$ \sqrt k \left( - \frac{1}{4 k^2} \right) < \sqrt k - 2 \sqrt {k+1} + \sqrt {k+2} < \sqrt k \left( - \frac{1}{4 k^2} + \frac{3}{8 k^3} \right), $$ $$ - \frac{1}{4 k^{3/2}} < \sqrt k - 2 \sqrt {k+1} + \sqrt {k+2} < - \frac{1}{4 k^{3/2}} + \frac{3}{8 k^{5/2}} \; \; . $$ When $k \geq 2,$ we get $$ - \frac{1}{4 k^{3/2}} < \sqrt k - 2 \sqrt {k+1} + \sqrt {k+2} < 0 \; \; . $$

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$$\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2} = \color{red}{\sqrt{k + 2} - \sqrt{k + 1}} - \big(\color{blue}{\sqrt{k + 1} - \sqrt{k}}\big)$$

By Telescoping sum we get,

$$ \sum^n_{k=1}\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2}= =\sqrt{n + 2}-\sqrt{2} -(\sqrt{n+1} -1) =\\=\frac{1}{\sqrt{n + 2}+\sqrt{n + 1 }}+1-\sqrt{ 2}$$

$$ \sum^{\infty}_{k=1}\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2} =\lim_{n\to\infty}\frac{1}{\sqrt{n + 2}+\sqrt{n + 1 }}+1-\sqrt{ 2}= \color{blue}{1-\sqrt{ 2}}$$

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  • $\begingroup$ what about the middle terms such as $-\sqrt{3},-\sqrt{4},-\sqrt{5},..$ etc. as I can see they dont disapper $\endgroup$
    – daulomb
    Dec 7, 2017 at 20:25
  • $\begingroup$ @daulomb all the term are included in that why do yoibwant to somewhere else? $\endgroup$
    – Guy Fsone
    Dec 7, 2017 at 22:34

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