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

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.

• Forget the fancy denominators, replacing them by $n$ doesn't change the limit. It's a Rieman sum for the integral $$\int^1_0a^x\,dx=\frac{a-1}{\ln a}.$$ – Professor Vector Sep 7 '17 at 9:57
• If you want to be a bit more rigorous, notice that we have, $$\sum_{k=1}^n\frac{a^{k/n}}{n+1}\leq\sum_{k=1}^n\frac{a^{k/n}}{n+\frac 1k}\leq\sum_{k=1}^n\frac{a^{k/n}}n$$ where the LHS of the above inequality can be seen as the left Riemann sum of $\int_0^1a^x\,\mathrm dx$ where the interval is partitioned by $n+1$ points and the RHS as the standard right Riemann sum of the same integral. Now, apply squeeze theorem. – Prasun Biswas Sep 7 '17 at 10:12

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

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

• Very nice and direct +1 – Paramanand Singh Sep 7 '17 at 12:06