Find asymptotic of sequence $a_n=\sqrt[4]{1}+\sqrt[4]{2}+...+\sqrt[4]{n}, n \to \infty$.
Here is my solution:
$$\sqrt[4]{1}+\sqrt[4]{2}+...+\sqrt[4]{n}=n^\frac{5}{4}\sum\limits_{i=1}^{n} \sqrt[4]\frac{i}{n}\frac{1}{n} =n^\frac{5}{4}\left(\int\limits_{0}^{1} \sqrt[4]{x}dx+o(1)\right)=n^\frac{5}{4}\left(\frac{4}{5}+o(1)\right)$$
But I'm not sure that it's correct. Could someone please check it and let me know what is wrong?
It is correct, and the problem can be generalized.
The sum for a general power $p$ can be expressed by the generalized harmonic number as follows
$$s_{n}(p)=\sum _{k=1}^n k^p = H_n^{(-p)}$$
The asymptotic behaviour of the harmonic number is then (for $p \ne -1$)
$$H_n^{(-p)} \to \left(\frac{n^{p+1}}{p+1}+\frac{1}{2}n^p\right)+\zeta (-p)+\frac{p}{12 n} +O(\frac{1}{n^3})$$
and for $p=-1$
$$H_n^{(1)}=H_n \to \log (n)+\gamma+\frac{1}{2 n}-\frac{1}{12 n^2}+O(\frac{1}{n^3})$$
Hence the series is divergent for $p\ge-1$ and convergent otherwise.