Find the infinite sum $\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} $ 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.
 A: Hint: $$\sum _{ k=1 }^{ \infty  } \sqrt { k } -2\sqrt { k+1 } +\sqrt { k+2 } =\sum _{ k=1 }^{ \infty  } \left( \sqrt { k+2 } -\sqrt { k+1 }  \right) +\sum _{ k=1 }^{ \infty  } \left( \sqrt { k } -\sqrt { k+1 }  \right) \\  $$
A: Hint: use
$$
\sum_{k=0}^\infty (b_{k+1}-b_k)=\lim_{N\to\infty} b_N - b_1
$$
and choose $b_k$ wisely.
A: 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}
$.
