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I am facing a problem, where I have to find the partial sum of a sequence/sum and with that, the infinite sum of the sequence.

$\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} $

The problem here is that I don't know how to proceed. I would be thankful if someone would steer me in the right direction.

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    This sum telescopes. – 2017-02-01
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    Sorry about that. I copied the sum text from another question, forgot to change the n to k. Changed it now. – 2017-02-01
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    Yes was gonna say telescope but some fast guy already scoped it. – 2017-02-01
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    See also http://math.stackexchange.com/questions/124708/finding-a-minorant-to-sqrtk1-sqrtk and http://math.stackexchange.com/questions/1661080/evaluate-sum-sqrtn1-sqrt-n and http://math.stackexchange.com/questions/883457/why-does-sum-n-1-infty-sqrtn1-sqrtn-diverge Found [using Approach0](https://approach0.xyz/search/?q=%24%5Csum%20_%7B%20k%3D1%20%7D%5E%7B%5Cinfty%7D%20%5Cleft(%5Csqrt%20%7Bk%2B1%20%7D%20-%5Csqrt%20%7B%20k%20%7D%5Cright)%24&p=1). – 2017-02-01

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Hint: $$\sum _{ k=1 }^{ \infty } \sqrt { k } -2\sqrt { k+1 } +\sqrt { k+2 } =\sum _{ k=1 }^{ \infty } \left( \sqrt { k+2 } -\sqrt { k+1 } \right) +\sum _{ k=1 }^{ \infty } \left( \sqrt { k } -\sqrt { k+1 } \right) \\ $$

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    I think I'm taking a wrong route somewhere. I proceeded to check some of the K values. this scoped down to $1-\sqrt {2}-\sqrt {k+1} + \sqrt {k+2}$ but that's where I ran out of ideas. Am I doing something wrong? – 2017-02-01
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    @Janno Can you prove that $\sqrt{k+2}-\sqrt{k+1}$ tends to 0 as $k\to\infty$? – 2017-02-01
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    Yes, which leaves me with $1-\sqrt 2$, yet wolfram insists the answer is -1. That far I got when I started giving k values to the initial sum sequence as well. But the reason that Wolfram shows the answer as -1, is what brought me here. [Link to Wolfram](https://www.wolframalpha.com/input/?i=sum(sqrt(k)-2*(sqrt(k%2B1))%2Bsqrt(k%2B2))) – 2017-02-01
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Hint: use $$ \sum_{k=0}^\infty (b_{k+1}-b_k)=\lim_{N\to\infty} b_N - b_1 $$ and choose $b_k$ wisely.

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You want $\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} $.

In general,

$\begin{array}\\ \sum_{k=1}^{n}(f(k)-2f(k+1)+f(k+2)) &=\sum_{k=1}^{n}f(k)-2\sum_{k=1}^{n}f(k+1)+\sum_{k=1}^{n}f(k+2)\\ &=\sum_{k=1}^{n}f(k)-2\sum_{k=2}^{n+1}f(k)+\sum_{k=3}^{n+2}f(k)\\ &=(f(1)+f(2)+\sum_{k=3}^{n}f(k))\\ &\quad -2(f(2)+f(n+1)+\sum_{k=3}^{n}f(k))\\ &\quad+(f(n+1)+f(n+2)+\sum_{k=3}^{n}f(k))\\ &=(f(1)+f(2))-2(f(2)+f(n+1))+(f(n+1)+f(n+2))\\ &=f(1)-f(2)+(f(n+2)-f(n+1))\\ \end{array} $

Therefore, if $\lim_{n \to \infty} (f(n+2)-f(n+1)) = 0$, then $\sum_{k=1}^{\infty}(f(k)-2f(k+1)+f(k+2)) = f(1)-f(2) $.

If $f(k) = \sqrt{k}$,

$\begin{array}\\ f(n+2)-f(n+1) &=\sqrt{n+2}-\sqrt{n+1}\\ &=(\sqrt{n+2}-\sqrt{n+1})\dfrac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt{n+2}+\sqrt{n+1}}\\ &=\dfrac{1}{\sqrt{n+2}+\sqrt{n+1}}\\ &\lt \dfrac1{2\sqrt{n}}\\ & \to 0 \text{ as } n \to \infty\\ \end{array} $

so the sum is $f(1)-f(2) =1-\sqrt{2} $.

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    Thank you for your input, I reached the same answer after I wrote out the K values and saw that the items start to cancel out, but the problem arised with Wolfram which insisted that the answer is -1 instead of $1 - \sqrt 2$. – 2017-02-02