# 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 $\infty$.
Looking at the expression you can conclude that it is an infinite series. $$\sum_{n=1}^\infty \frac{4n}{\sqrt[5]{n^{10}+n}}$$. Taking the convergence test for the sequence you can arrive at: $$\lim_{n \rightarrow \infty}\frac{4n}{\sqrt[5]{n^{10}+n}}$$. You can then divide by n in both numerator and denominator and assuring that you raise n to the 5th power and take the 5th root in the denominator so you can get n inside the root. The expression becomes: $$\lim_{n \rightarrow \infty}\frac{4}{\sqrt[5]{n^{5}+\frac{1}{n^{4}}}}=0$$. Therefore you may conclude that the series converges as it has passed the convergence test, however it turns out that this infinite series behaves just as the harmonic series and therefore it diverges.