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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!

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    $\begingroup$ This problem is not well-defined. Take $n=0$. Are we allowing complex-valued summands? $\endgroup$ – user28877 Nov 20 '13 at 22:31
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    $\begingroup$ This is just a telescoping series. So what do you get as a result? Then just turn this into rigour. $\endgroup$ – Chris K Nov 20 '13 at 22:34
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    $\begingroup$ @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. $\endgroup$ – Chris K Nov 20 '13 at 22:41
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    $\begingroup$ It's telescoping. I don't see why anyone would approach the problem in a more complicated way. $\endgroup$ – Doc Nov 20 '13 at 22:49
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    $\begingroup$ A correction to my earlier comment... should be $\sqrt{N+1}$ and not $\sqrt{N}$ but the idea stays the same. $\endgroup$ – Chris K Nov 20 '13 at 22:53
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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.

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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.

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