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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.

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    $\begingroup$ Forget the fancy denominators, replacing them by $n$ doesn't change the limit. It's a Rieman sum for the integral $$\int^1_0a^x\,dx=\frac{a-1}{\ln a}.$$ $\endgroup$ – Professor Vector Sep 7 '17 at 9:57
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    $\begingroup$ 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. $\endgroup$ – Prasun Biswas Sep 7 '17 at 10:12
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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 $$ \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+\frac12}+\frac{\sqrt[n]{a^3}}{n+\frac13}+\cdots+\frac{\sqrt[n]{a^n}}{n+\frac1n}\right) \geq \left(\frac{\sqrt[n]a}{n+1}+\frac{\sqrt[n]{a^2}}{n+1}+\frac{\sqrt[n]{a^3}}{n+1}+\cdots+\frac{\sqrt[n]{a^n}}{n+1}\right)=\sqrt[n]{a}.\frac{a-1}{(n+1)\left(\sqrt[n]{a}-1\right)} $$

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

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

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