$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)=?$

Question

What is the value of limit

$$\lim_{n \to \infty}\left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right)$$

If we know that $a>0$?

I get stuck on this, it seems to be Riemann sum but I can't find relation. I am thankful if someone could guide me.

Answer 1

Hint:

$$ \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \leq \left( \frac{\sqrt[n]a}{n}+\frac{\sqrt[n]{a^2}}{n}+\frac{\sqrt[n]{a^3}}{n}+\cdots+\frac{\sqrt[n]{a^n}}{n}\right)=\sqrt[n]{a}.\frac{a-1}{n\left(\sqrt[n]{a}-1\right)} $$ and $$ \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \geq \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+1}+\frac{\sqrt[n]{a^3}}{n+1}+\cdots+\frac{\sqrt[n]{a^n}}{n+1}\right)=\sqrt[n]{a}.\frac{a-1}{(n+1)\left(\sqrt[n]{a}-1\right)} $$

$\left(\text{ The answer would be }\dfrac{a-1}{\log a}.\right)$