# Determine the convergence of $\sum\limits_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2}$ [duplicate]

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.

• It telescopes${}$. Dec 7, 2017 at 19:55
• 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? Dec 7, 2017 at 20:06
• @John.Doh: $$\lim_{n\to +\infty}\left[\sqrt{1}-\sqrt{2}+\sqrt{n+2}-\sqrt{n+1}\right]$$ is finite. Dec 7, 2017 at 20:18
• 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. Dec 7, 2017 at 20:29

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.

• but isn't it $$\sqrt{k} - 2 \sqrt{k + 1} + \sqrt{k + 2} = \sqrt{k + 2} - \sqrt{k + 1} - \big(-\sqrt{k} + \sqrt{k + 1}\big)$$ ? Dec 7, 2017 at 20:08
• Yes, a typo. The point is the same, though.
– user296602
Dec 7, 2017 at 20:09
• @user296602: but the derivative of root $x$ is $\frac{1}{2x^{1/2}}$. Dec 7, 2017 at 20:17
• @daulomb Yes, but this is a second difference, which is related to the second derivative.
– user296602
Dec 7, 2017 at 20:35

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 $$- \frac{1}{4 k^{3/2}} < \sqrt k - 2 \sqrt {k+1} + \sqrt {k+2} < 0 \; \; .$$

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

By Telescoping sum we get,

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

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

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