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Im trying to check if the series

$\sum_{k=2}^{\infty} \frac{1}{\sqrt{k-1}} - \frac{1}{\sqrt{k+1}} $

is converging or not.

I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.

For the series with even k >=2 I have started writing down the terms:

$S_k = (1-\frac{1}{\sqrt{3}})+ (\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}})+(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}})+...$

But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.

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    $\begingroup$ Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. ! $\endgroup$ – Matti P. Apr 3 '19 at 8:57
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Since$$\sum_{k=2}^\infty\left(\frac1{\sqrt{k-1}}-\frac1{\sqrt{k+1}}\right)=\sum_{k=2}^\infty\left(\frac1{\sqrt{k-1}}-\frac1{\sqrt k}\right)+\sum_{k=2}^\infty\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right),$$the sum of your series is $1+\dfrac1{\sqrt2}$.

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  • $\begingroup$ Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…) $\endgroup$ – Evargalo Apr 3 '19 at 14:31
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    $\begingroup$ @Evargalo I agree that it is a good example for that discussion. $\endgroup$ – José Carlos Santos Apr 3 '19 at 14:41
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To remain on the safe side of strictness consider the partial sum

$$s(n) = \sum_{k=2}^n \left(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k+1}}\right)$$

and then look for the limit $n\to\infty$.

We have

$$s(n) = \left( \frac{1}{\sqrt{1}} +\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+ ...+ \frac{1}{\sqrt{n-1}}\right)\\ -\left( \frac{1}{\sqrt{3}} +\frac{1}{\sqrt{4}}+ ...+ \frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\\ = 1+\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} $$

and we get finally

$$\lim_{n\to \infty } \, s(n) =1+\frac{1}{\sqrt{2}} $$

EDIT

The generalization to an arbitrary sequence $\{a(k)\}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by

$$s(d,n) = \sum_{k=1}^n \left( a_{n}-a_{k+d} \right)= \sum_{j=1}^d a_j-\sum_{j=1}^d a_{n+j}$$

And if $\lim_{n\to \infty } \, a_n =0$ we find

$$s(d) = \lim_{n\to \infty } \, s(d,n) =\sum_{j=1}^d a_j$$

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