Define for $n\in\mathbb{N}$ $$S_n=\sum_{r=0}^{n}\binom{n}{r}^2\left(\sum_{k=1}^{n+r}\frac{1}{k^5}\right)$$
I need to find $\{S_n\}$ for $n$ large where $\{x\}$ denotes the fractional part of $x$.
$$S_n=\sum_{r=0}^{n}\binom{n}{r}^2\left(1+\sum_{k=2}^{n+r}\frac{1}{k^5}\right)$$ So we get $$S_n=\sum_{r=0}^{n}\binom{n}{r}^2+\sum_{r=0}^{n}\binom{n}{r}^2\left(\sum_{k=2}^{n+r}\frac{1}{k^5}\right)$$ Now we have by Wolfram Alpha $$S_n=\binom{2n}{n}+\sum_{r=0}^{n}\binom{n}{r}^2\left(\sum_{k=2}^{n+r}\frac{1}{k^5}\right)$$ Now since $\binom{2n}{n}\in\mathbb{N}$, so we have $$\{S_n\}=\left\{\sum_{r=0}^{n}\binom{n}{r}^2\left(\sum_{k=2}^{n+r}\frac{1}{k^5}\right)\right\}$$ We obtain $$\{S_n\}=\left\{\sum_{r=0}^{n}\binom{n}{r}^2\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{(n+r)^5}\right)\right\}$$ Hence $$\{S_n\}=\left\{\binom{n}{0}^2\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{n^5}\right)+\binom{n}{1}^2\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{(n+1)^5}\right)+...+\binom{n}{n}^2\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{(2n)^5}\right)\right\}$$ So we get $$\{S_n\}=\left\{\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{n^5}\right)\sum_{r=0}^{n}\binom{n}{r}^2+\frac{\binom{n}{1}^2}{(n+1)^5}+\binom{n}{2}^2\left(\frac{1}{(n+1)^5}+\frac{1}{(n+2)^5}\right)+...+\binom{n}{n}^2\left(\frac{1}{(n+1)^5}+\frac{1}{(n+2)^5}+...+\frac{1}{(2n)^5}\right)\right\}$$ Now we get $$\{S_n\}=\left\{\binom{2n}{n}\left(\frac{1}{2^5}+\frac{1}{3^5}+...+\frac{1}{n^5}\right)+\frac{1}{(n+1)^5}\sum_{r=1}^{n}\binom{n}{r}^2+\frac{1}{(n+2)^5}\sum_{r=2}^{n}\binom{n}{r}^2+...+\frac{1}{(2n)^5}\binom{n}{n}^2\right\}$$ Therefore $$\{S_n\}=\left\{\binom{2n}{n}\sum_{k=2}^{n}\frac{1}{k^5}+\frac{1}{(n+1)^5}\left(\binom{2n}{n}-\binom{n}{0}^2\right)+\frac{1}{(n+2)^5}\left(\binom{2n}{n}-\binom{n}{0}^2-\binom{n}{1}^2\right)+...+\frac{1}{(2n)^5}\binom{n}{n}^2\right\}$$ Hence $$\{S_n\}=\left\{\binom{2n}{n}\sum_{k=2}^{2n}\frac{1}{k^5}+\sum_{k=1}^{n}\frac{1}{(n+k)^5}\left(\binom{2n}{n}-\sum_{r=0}^{k-1}\binom{n}{r}^2\right)\right\}$$ Any help would be highly appreciated. Thanks!
