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For $\sqrt{(n + 1)} - \sqrt{n}$, sequence I am testing convergence and limits.
Indeed it seems converging. Then,
$$\lim_{n\to \infty} \sqrt{(n + 1)} - \sqrt{n} = 0 $$ But if telescopic series taken,
$$\lim_{N\to \infty} \sum_{n=1}^N\left[\sqrt{(n + 1)} - \sqrt{n}\right]$$ $$i.e. \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{N-1+1}-\sqrt{N-1}+\sqrt{N+1}-\sqrt{N} $$

sums left with $$\lim_{N\to \infty} [-\sqrt{1}-\sqrt{N-1} + \sqrt{N+1} ] = -1$$.
I have tried so far this. Is this will be a good way to deal with sequence?

step-by-step
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    But if telescopic series taken Series of what? The previous $\lim$ referred to some sequence $\sqrt{(n + 1)} - \sqrt{n}$ which does indeed converge to $0$. Where does the series come into the picture, or otherwise put, what is your question? – dxiv Mar 18 '17 at 06:59
  • I think you made a mistake in your calculation. The term $-\sqrt(N-1)$ should be cancelled out. – Jerry Mar 18 '17 at 07:03
  • @dxiv I am looking for infinite sum of sequence from two ways. But not sure about can go this way or not. Thanks. – step-by-step Mar 18 '17 at 07:49
  • @Jerry, yup its mistake. Thanks. – step-by-step Mar 18 '17 at 07:49

2 Answers2

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In the telescoping sum you wrote, $$\lim_{N\to \infty} \sum_{n=1}^N\left[\sqrt{(n + 1)} - \sqrt{n}\right]$$

We can rearrange the terms so that we have the sum $$\lim_{N\to \infty}(-\sqrt1 - \sqrt2 - ... -\sqrt{N} + \sqrt2 + \sqrt3 +...+\sqrt{N} + \sqrt{N+1})$$

Once we cancel like terms, we have $\lim_{N \to \infty} (\sqrt{N+1} - \sqrt{1}) = \lim_{N \to \infty} (\sqrt{N+1} - 1) = \infty$

What this tells us is, even though the difference between two consecutive terms eventually reaches $0$, the sum of the consecutive differences need not converge.

Yunus Syed
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Hint: $$ \sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}} $$

robjohn
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