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or divergent??

I tried few tests, but I didn't success to discover if the series is convergent or is divergent...

$$\sum_{n=0}^{\infty} \sqrt{n+1}-\sqrt{n}$$

Thank you!

CS1
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    This problem is not well-defined. Take $n=0$. Are we allowing complex-valued summands? –  Nov 20 '13 at 22:31
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    This is just a telescoping series. So what do you get as a result? Then just turn this into rigour. – Christopher K Nov 20 '13 at 22:34
  • Sorry there was a typing mistake. I fix it! – CS1 Nov 20 '13 at 22:36
  • @ChrisK - can you help me now? – CS1 Nov 20 '13 at 22:37
  • I see a divergent series based on comparison with $\sum\limits_{n=1}^\infty \frac 1{\sqrt {2n}}$... – abiessu Nov 20 '13 at 22:37
  • @abiessu - How do you see it? – CS1 Nov 20 '13 at 22:37
  • If we take the conjugate ($\frac 1{\sqrt{n+1}+\sqrt n}=\sqrt {n+1}-\sqrt n$) then the comparison is clear. – abiessu Nov 20 '13 at 22:39
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    @YoavFridman, when you take the sum from $0$ to $N$, you get $\sqrt{N} - \sqrt{0} = \sqrt{N}$. Now we can always take $k$ sufficiently large such that $|\sqrt{N} - \sqrt{k}|>\varepsilon$ where we let $\varepsilon>0$. So, if you treat the series like a sequence, it is not Cauchy. So, by contrapositive, the series can't be convergent. – Christopher K Nov 20 '13 at 22:41
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    It's telescoping. I don't see why anyone would approach the problem in a more complicated way. – Doc Nov 20 '13 at 22:49
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    A correction to my earlier comment... should be $\sqrt{N+1}$ and not $\sqrt{N}$ but the idea stays the same. – Christopher K Nov 20 '13 at 22:53
  • @YoavFridman : It's a telescoping series. – Stefan Smith Nov 20 '13 at 23:07

2 Answers2

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Let $S_n$ be the sequence of partial sums: $$S_n = \sum_{k=0}^n \sqrt{k+1} - \sqrt{k}$$ It is easy to see that $S_0 = 1$ and by telescoping $$S_n = \sqrt{n+1}$$ Since convergence of a series is defined through convergence of the partial sums and since $S_n$ obviously diverges, the series diverges as well.

AlexR
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First, note that

$$ a_n=\sqrt{n+1}-\sqrt{n}=\frac{1}{ \sqrt{n+1}+\sqrt{n}}\sim_{n\to \infty} \frac{1}{2\sqrt{n}}=b_n $$

Now, use the following result and see what you get.

Community
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Mhenni Benghorbal
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