Evaluation of limit with summation till infinity. The question to be solved is:
$$ \lim_{n \to \infty} \left( \ \sum_{k=10}^{n+9} \frac{2^{11(k-9)/n}}{\log_2 e^{n/11}}  \ - \sum_{k=0}^{n-1} \frac{58}{\pi\sqrt{(n-k)(n+k)}} \ \right)$$
The first thing that occured to me was to transform the limits into definite integrals using the limit definition of integrals, so it'll become easier to evaluate.
However, I have no clue how to convert them into definite integrals. Could anyone please shed some light on how to proceed? Or is there a better way to solve this problem?
 A: $$ \lim_{n \to \infty} \left( \ \sum_{k=10}^{n+9} \frac{2^{11(k-9)/n}}{\log_2 e^{n/11}}  \ - \sum_{k=0}^{n-1} \frac{58}{\pi\sqrt{(n-k)(n+k)}} \ \right)$$
By putting $k=m+9$ in the first sum, we get
\begin{align}
\sum_{k=10}^{n+9} \frac{2^{11(k-9)/n}}{\log_2 e^{n/11}}
&=\frac 1{\log_2 e^{n/11}}\sum_{m=1}^{n}(2^{11/n})^m\\
&=\frac 1{\log_2 e^{n/11}}\frac{(2^{11/n})^{n+1}-2^{11/n}}{2^{11/n}-1}\\
&=\frac{11\log(2)}{n}\frac{2^{11/n}(2^{11}-1)}{2^{11/n}-1}\\
&=\frac{11\log(2)}{n}\frac{2^{11/n}(2^{11}-1)}{e^{11/n\log(2)}-1}\\
&\sim\frac{11\log(2)}{n}\frac{2^{11}-1}{11/n\log(2)}\\
&\xrightarrow{n\to\infty}2^{11}-1
\end{align}
For the second sum
\begin{align}
\sum_{k=0}^{n-1} \frac{58}{\pi\sqrt{(n-k)(n+k)}}
&=\frac{58}\pi\sum_{k=0}^{n-1} \frac 1{\sqrt{1-(\frac kn)^2}}\frac 1n\\
&\xrightarrow{n\to\infty}\frac{58}\pi\int_0^1\frac{\mathrm dx}{\sqrt{1-x^2}}\\
&=\frac{58}\pi\frac\pi 2\\
&= 29
\end{align}
A: For converting an sum to definite integral :
Convert your sum having limits $a$ to $b$ to the form $f(k/n) * 1/n$.
Take $k/n$ as $x$, $1/n$ as $dx$ and set the limits as $\frac{a}{n}$ and $\frac{b}{n}$.  This is actually the sum of all tiny rectangles (vertical ones) when we break the area under a curve into small rectangles of width $\frac{1}{n}$.
For the first term, simplify the denominator and take $k-9=t$ . The first term becomes $2^{11\frac{t}{n}}$ $\frac{11\ln2}{n}$. You can now convert it to an Definite Integral and evaluate it. (It comes out to be $2^{11} - 2^0$).
For the second term, take 'n' out of the denominator and convert it to an integral to get $29$. 
The answer comes out to be 'this year'.
A: The first summation is essentially a geometric series, and with $j:=k+9$,
$$\frac{2^{11(k-9)/n}}{\log_2 e^{n/11}}=\frac{\left(2^{11/n}\right)^j}{\dfrac{n}{11}\log_2e}.$$
Then summing from $1$ to $n$,
$$\frac{2^{11/n}(2^{11}-1)}{\dfrac n{11}\log_2e(2^{11/n}-1)}.$$
As 
$$2^{11/n}-1=e^{11\ln2/n}-1=1+\frac{11\ln2}{ n}+\cdots-1,$$ the limit reduces to $2^{11}-1$.

As an integral, 
$$\frac{2^{11(k-9)/n}}{\log_2 e^{n/11}}\to 11\ln2\,2^{11x}dx$$ which integrates as $2^{11x}$, from $0$ to $1$.
