Find the sum of series: $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$

Question

Find the sum of series:

$$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$$

My Attempt:

I tried to go by telescopic method but nothing appears to be cancelling.

Something similar was given in a book by Titu Andreescu. I will try to reproduce

Let $S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$

Further let $T=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$

Clearly $S+T=\sqrt{100}-1=9$

Also $S-T=2S+1-\sqrt{100}$

and $S>T$

$\Rightarrow 2S>S+T$

$\Rightarrow S>4.5$

This is the best I could come up with

Answer 1

Let $S_m$ denote $\sum_{j=1}^m\sqrt{j}$. (In terms of Hurwitz eta functions $S_n=\zeta(-\frac12)-\zeta(-\frac12,\,m+1)$.) Take $n=50$ in $$\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k}}=\sum_{k=1}^n(\sqrt{2k}-\sqrt{2k-1})=\sqrt{2}\sum_{k=1}^n\sqrt{k}-\sum_{2|k,\,1\le k\le 2n}\sqrt{k}\\=2\sqrt{2}\sum_{k=1}^n\sqrt{k}-\sum_{k=1}^{2n}\sqrt{k}=2\sqrt{2}S_n-S_{2n}.$$I think that's about the best we can do.

Answer 2

I don't think it's possible to say for sure, but you can find the whole part
Hint: Try to find $S - T$.

$$S - T = \dfrac{1}{\sqrt 2 + 1} - \left(\dfrac{1}{\sqrt 2 + \sqrt 3} - \dfrac{1}{\sqrt 3 + \sqrt 4}\right) - . . . - \left(\dfrac{1}{\sqrt 98 + \sqrt 99} - \dfrac{1}{\sqrt 99 + \sqrt 100}\right) < \dfrac{1}{\sqrt 2 + 1} < 1$$ since each bracket is greater than $0$
$S - T < 1$ and $S + T = 9$
So if the whole part is $S > 5$, then $T > 4$, which means $S + T > 9$, a contradiction $\lfloor {S} \rfloor = 5$ and $\{ {S} \}$ < $\dfrac{1}{\sqrt 2 + 1}$