<s>[INST] calculating a limit of sequence Can some one help to show that for $a>0$
$$ \lim_{n\to \infty} \frac{a^\frac{1}{n}}{n+1}+\frac{a^\frac{2}{n}}{n+\frac{1}{2}}+...+\frac{a^\frac{n}{n}}{n+\frac{1}{n}}=\frac{a-1}{\log a}$$
My 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.
 [/INST] 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, 
$$\lim_{n\to \infty} S(n) = \frac{a - 1}{\log a}.$$
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