Can some one help to show that for a>0

limMy tries : I was not able to use riemann sum for calculating.

For the special case a=1 the RHS has limit 1 and for the LHS for a=1 we can write\frac{n}{n+1} \le \frac{1}{n+1}+\frac{1}{n+\frac{1}{2}}+…+\frac{1}{n+\frac{1}{n}} \le \frac{n}{n+\frac{1}{n}} so the limit of LHS is 1 . For arbitrary a I have failed to answer.

**Answer**

Let

S(n) = \frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + \frac{1}{2}} + \cdots + \frac{a^{\frac{n}{n}}}{n + \frac{1}{n}}.

Then

\frac{a^{\frac{1}{n}}}{n + 1} + \frac{a^{\frac{2}{n}}}{n + 1} + \cdots + \frac{a^{\frac{n}{n}}}{n + 1} < S(n) < \frac{a^{\frac{1}{n}}}{n} + \frac{a^{\frac{2}{n}}}{n} +\cdots + \frac{a^{\frac{n}{n}}}{n}

that is,

\frac{n}{n+1}\cdot \frac{a^{\frac{1}{n}} + a^{\frac{2}{n}} + \cdots + a^{\frac{n}{n}}}{n} < S(n) < \frac{a^{\frac{1}{n}} + a^{\frac{2}{n}}+\cdots + a^{\frac{n}{n}}}{n}. \tag{*}

The left- and right-most sides of (*) converge to

\int_0^1 a^x\, dx = \frac{a^x}{\log a}\bigg|_{x = 0}^1 = \frac{a - 1}{\log a}.

Hence, by the squeeze theorem,

\lim_{n\to \infty} S(n) = \frac{a - 1}{\log a}.

**Attribution***Source : Link , Question Author : Fin8ish , Answer Author : kobe*