<s>[INST] Find the sum of series: $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$ 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
 [/INST] 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.
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