Squeeze theorem: sequence of sums So I've been having some problems with finding the limit of this sequence:
 $$\lim_{n \rightarrow \infty}\left(\frac{4}{ \sqrt[5]{1^{10}+1 }} + \frac{8}{\sqrt[5]{2^{10}+2}} + ... + \frac{4n}{\sqrt[5]{n^{10}+n}}\right)$$
I've tried to find it using the squeeze theorem, here's what I've tried:
 $$ \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}+n}} \leq  \frac{4}{ \sqrt[5]{1^{10}+1 }} + \frac{8}{\sqrt[5]{2^{10}+2}} + ... + \frac{4n}{\sqrt[5]{n^{10}+n}} \leq \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}}} $$
and:
 $$\lim_{n \rightarrow \infty} \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}+n}} = \lim_{n \rightarrow \infty} \frac{\frac{4+4n}{2}\cdot n }{\sqrt[5]{n^{10}+n}} = \lim_{n \rightarrow \infty} \frac{2n^{2}+2}{\sqrt[5]{n^{10}+n}} = 2 $$
$$\lim_{n \rightarrow \infty} \frac{4 + 8 + ... + 4n}{\sqrt[5]{n^{10}}} = \lim_{n \rightarrow \infty} \frac{\frac{4+4n}{2}\cdot n }{\sqrt[5]{n^{10}}} = \lim_{n \rightarrow \infty} \frac{2n^{2}+2}{\sqrt[5]{n^{10}}} = 2 $$
Therefore the limit of my initial sequence should also be 2. But I find it highly unlikely, mostly because:
$$ \sum_{n=1}^{\infty} \frac{4n}{\sqrt[5]{n^{10}+n}} \text{ diverges}$$ 
and:
$$ \frac{4}{\sqrt[5]{1^{10}+1}} \approx 3.482202253184496556545...$$ 
So I must have chosen the wrong sequences to bound my original one. Where did I go wrong?

Consider $$s_n=\frac{4 n}{\sqrt[5]{n^{10}+n}}=\frac{4 n}{n^2\sqrt[5]{1+\frac{1}{n^9}}}=\frac{4}{n} \frac{1}{\sqrt[5]{1+\frac{1}{n^9}}}$$ So, for large values of $n$, it write $$s_n=\frac{4}{n}-\frac{4}{5 n^{10}}+\frac{12}{25
   n^{19}}+O\left(\left(\frac{1}{n}\right)^{21}\right)$$ and, so, the sum $\sum_{n=1}^{\infty}s_n$ looks very much to the harmonic number which goes to $\infty$.
