Show if the sequence $s_n :=\sum\limits _{j= 2^n}^{2^{n+1} }{ \frac 1 { j \sqrt [ n ]{ j } } } $ has a limit I'm still onto reviewing some Calculus 1.
and I want to show wether the sequence $$s_n :=\sum _{j=2^n}^{2^{n+1}} \frac 1 {j\sqrt[ n ]{j}}$$ has a limit or not.
Since $s_n =\sum\limits_{j=2^n}^{2^{n+1}} \frac 1 { j^{\frac 1 n +1} }$ 
looks like a geometric sum it probably has one but i couldnt find it yet.
 A: Simple upper and lower bounds allow to bypass any advanced knowledge. To wit, note that, for every $2^n\leqslant j\leqslant2^{n+1}$, $$2\leqslant\sqrt[n] j\leqslant2\cdot2^{1/n}$$ hence, for every $n$, $$\frac12\cdot t_n\leqslant s_n\leqslant\frac12\cdot\frac1{2^{1/n}}\cdot t_n$$ where $$t_n=\sum_{j=2^n}^{2^{n+1}}\frac1j=\frac1{2^n}+\frac1{2^n}\sum_{j=2^n+1}^{2^{n+1}}\frac{2^n}j$$ The last sum on the RHS is a Riemann sum of the function $f(x)=\frac1x$ on the interval $(1,2)$ hence $$\lim_{n\to\infty}t_n=0+\int_1^2f(x)dx=\ln2$$ and $$\lim_{n\to\infty}s_n=\frac12\cdot\ln2$$
A: Assuming that $n$ is a large natural number, the difference between 
$$ S_n=\sum_{j=2^n}^{2^{n+1}}\frac{1}{j^{1+\frac{1}{n}}}\qquad\text{and}\qquad I_n=\int_{2^n}^{2^{n+1}}\frac{dx}{x^{1+\frac{1}{n}}}$$
is negligible. Indeed, by invoking the Hermite-Hadamard inequality we have that such difference is bounded by $\frac{4}{2^n}$. $I_n$ can be computed in a explicit way and the problem boils down to computing:
$$ \lim_{n\to +\infty} n\left(\frac{1}{2}-\frac{1}{2^{1+\frac{1}{n}}}\right).$$
The Taylor series of $2^{-1-x}$ at the origin is given by $\frac{1}{2}-\frac{\log 2}{2}x+O(x^2)$, hence the previous limit equals $\color{red}{\frac{\log 2}{2}}\approx 0.34657359$.
