# 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?

• The first term is easy to compute. Just focus on the second as a Riemann sum. – Claude Leibovici Oct 2 '18 at 8:18

$$\lim_{n \to \infty} \left( \ \sum_{k=10}^{n+9} \frac{2^{11(k-9)/n}}{\log_2 e^{n/11}} \ - \sum_{k=0}^{n-1} \frac{58}{\pi\sqrt{(n-k)(n+k)}} \ \right)$$

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}

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'.

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 $$1$$.