Find the infinite sum $\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} $

Question

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.

Answer 1

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

Answer 2

Hint: use $$ \sum_{k=0}^\infty (b_{k+1}-b_k)=\lim_{N\to\infty} b_N - b_1 $$ and choose $b_k$ wisely.

Answer 3

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} $.