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.

  • 1
    $\begingroup$ 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

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 $1+\dfrac1{\sqrt2}$.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @Evargalo I agree that it is a good example for that discussion. $\endgroup$ – José Carlos Santos Apr 3 '19 at 14:41

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

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


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

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.