find $\lim_n\frac{a^\frac1n}{n+1}+\frac{a^\frac 2n}{n+\frac 12}+...+\frac{a^\frac nn}{n+\frac 1n}$ using Riemann integral Here is the question: 
prove that $S_n=\frac{a^\frac1n}{n+1}+\frac{a^\frac 2n}{n+\frac 12}+...+\frac{a^\frac nn}{n+\frac 1n}$ is convergent for $a>0$ then find its limit.
My attempt: If we accept that $S_n$ is convergent then by multiplying each sides by $a^\frac1n$ have:
$$\lim_{n\to\infty}n(a^\frac 1n-1)S_n=n\frac{-a^\frac 1n}{n+1}+na^\frac 2n(\frac{1}{n+1}-\frac{1}{n+\frac 12})+...+na^\frac nn(\frac1{n+\frac 1n}-\frac 1{n+\frac 1{n-1}})+n\frac{a^\frac{n+1}n}{n+\frac 1n}$$
so $$\lim S_n=\frac{a-1}{\ln a}$$
(I think we should assume $a\neq 1)$
However usually people calculate such limits using Riemann integral. I would like to use Riemann integral to find the value of the limit.
Thank you kindly for your help.

Edit:
$n\frac{-a^\frac 1n}{n+1}+n\frac{a^\frac{n+1}n}{n+\frac 1n}=n\frac{-a^\frac 1n}{n+1}+na^\frac 2n(\frac{1}{n+\frac 12}-\frac{1}{n+\frac 12})+...+na^\frac nn(\frac1{n+\frac 1{n-1}}-\frac 1{n+\frac 1{n-1}})+n\frac{a^\frac{n+1}n}{n+\frac 1n}\leq n(a^\frac1n-1)S_n\leq n\frac{-a^\frac 1n}{n+1}+na^\frac 2n(\frac{1}{n+1}-\frac{1}{n+1})+...+na^\frac nn(\frac1{n+\frac 1n}-\frac 1{n+\frac 1{n}})+n\frac{a^\frac{n+1}n}{n+\frac 1n}=n\frac{-a^\frac 1n}{n+1}+n\frac{a^\frac{n+1}n}{n+\frac 1n}$
 A: The sum
$$S_n = \sum_{n=1}^n \frac{a^{{i\over n}}}{n + \frac{1}{i}} = \sum_{i=1}^n \frac{1}{n}\frac{a^{{i\over n}}}{1 + \frac{1}{n^2}\frac{n}{i}}$$
is not quite a Riemann-sum since the summand is not on the form $\frac{1}{n}f\left({i\over n}\right)$ for some ($n$-independent) function $f$. However, it it very close to the Riemann-sum
$$\tilde{S}_n =  \sum_{i=1}^n \frac{1}{n}a^{{i\over n}}$$
which converges to the integral $\int_0^1a^{x}{\rm d}x$. The difference between the two sums $\tilde{S}_n$ and $S_n$ satisfy
$$0 \leq \tilde{S}_n - S_n = \sum_{i=1}^n \frac{a^{{i\over n}}}{n}\frac{1}{1 + ni} \leq \frac{1}{n}\tilde{S}_n$$
and since $\tilde{S}_n$ converges we have $\lim_\limits{n\to\infty} S_n - \tilde{S}_n = 0$ and it follows that
$$\lim\limits_{n\to\infty}\sum_{i=1}^n \frac{1}{n}\frac{a^{{i\over n}}}{n + \frac{1}{i}} = \int_0^1 a^x {\rm d}x = \frac{a-1}{\log(a)}$$
A: Well to convert to Riemann integral, you expect $i/n\to x$ and $1/n\to dx$. Then you just play around:
$$\sum_{i=1}^n \frac{a^{i/n}}{n+1/i}=\sum_{i=1}^n dx \frac{a^{x}}{1+1/(xn^2)}\to \int_0^1  a^x dx$$
What Riemann sum actually does is just limit the interval width $dx$ to zero, and covers the $x$ uniformly and continuously across the available range from $1/n$ to $n/n$. You start by isolating the $1/n$ term in front and check if the rest can be expressed with $i/n$.
The only mystery is the $1/(xn^2)$. This one you can just eliminate in the limit, because when $n\to \infty$, it is negligible compared to $1$ (it's actually better to leave it as $1/(in)$ because $in>n\to\infty$).
In the most standard textbook examples, there are no "extra" terms where $n$ is left standing alone. In that case, when you convert to $dx$ and $x$, just replace $\sum$ with $\int$. But here, there was something left, but fortunately went to zero in the limit.
Strictly speaking, we should not have done that without a proof (limit each term separately and then take the Riemann limit afterwards) but it worked out fine :)
