If $S = \sum_{k=1}^{99} \frac{(-1)^{k+1}}{\sqrt{k(k+1)}(\sqrt{k+1} - \sqrt{k})}$ , find $10S$

Question

If $$ S = \sum_{k=1}^{99} \frac{(-1)^{k+1}}{\sqrt{k(k+1})(\sqrt{k+1} - \sqrt{k})} , $$ find $10S.$

What I tried : Multiplying by the conjugate of the denominator I get: $$ S = \sum_{k=1}^{99} \frac{(-1)^{k+1}\sqrt{k(k+1)}(\sqrt{k+1} + \sqrt{k})}{k(k+1)}. $$

From here I thought this would be a telescopic sum and a partial fraction decomposition would take place. But unfortunately I didn't find any. This was my only thoughts on this problem so far.

Can anyone help?

Wolfram Alpha is giving the answer to be $S = 1.1$ , but how is it coming?

Answer 1

I think that the problem is in fact $$S = \sum_{k=1}^{99} \frac{(-1)^{k+1}}{\sqrt{k(k+1)}}\frac 1{\sqrt{k+1} - \sqrt{k} }$$ which now simplifies a lot since $$S= \sum_{k=1}^{99}(-1)^{k+1}\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\right)$$ And $10\,S$ is a whole number.

Edit

If we consider $$T_n= \sum_{k=1}^{n}(-1)^{k+1}\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\right)=1+\frac{(-1)^n}{\sqrt{2}}\left(\zeta \left(\frac{1}{2},\frac{n+3}{2}\right)-\zeta \left(\frac{1}{2},\frac{n+1}{2}\right)\right)$$ is seems that, for $1\leq n \leq 2^{30}$, there are only three values of $n$ such that $10\, S_n$ is a whole number (namely $3$, $24$ and $99$).

Answer 2

Please let me know if I miss something in the discussion, but I'm wondering why nobody calculated the sum explicitly yet: $$S= \sum_{k=1}^{n}(-1)^{k+1}\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\right) = \frac{(-1)^{n+1}}{\sqrt{n+1}}+1$$ after an index shift in the second sum $k+1\rightarrow k$.