# Undernetmath’s Weblog

### Solutions_2011

2011-10-16 Yet Another Limit by kmh
Determine $\displaystyle \lim_{n\to\infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k}$
Apply the Cesaro Stolz-Cesaro theorem:
$\displaystyle \lim_{n\to\infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k}=\lim_{n\to\infty} \frac{\displaystyle \sum_{k=1}^{n+1} \frac{2^k}{k} -\sum_{k=1}^n \frac{2^k}{k}}{\displaystyle \frac{2^{n+1}}{n+1}-\frac{2^n}{n}}=\lim_{n\to\infty} \frac{\displaystyle \frac{2^{n+1}}{n+1}}{\displaystyle \frac{2^n(n-1)}{n(n+1)}}=\lim_{n\to\infty} 2\frac{n}{n-1}=2$

Determine $\displaystyle \lim_{n\to \infty} \sum_{k=1}^n \frac{k^2}{(2k)^3+n^3}$
The basic idea is to recognize that the expression is a Riemann sum for a function of the type $\displaystyle \frac{f^\prime(x)}{f(x)}$.
$\displaystyle \lim_{n\to \infty} \sum_{k=1}^n \frac{k^2}{(2k)^3+n^3}=\lim_{n\to \infty} \frac{1}{24} \sum_{k=1}^n \frac{24k^2}{(2k)^3+n^3}=\lim_{n\to \infty} \frac{1}{24} \sum_{k=1}^n \frac{3\left(k\frac{2}{n}\right)^2}{\left(k\frac{2}{n}\right)^3+1}\cdot\frac{2}{n}$
$\displaystyle =\frac{1}{24} \int_0^2\frac{3x^2}{x^3+1}dx=\frac{1}{24}\left[\ln(x^3+1)\right]_0^2=\frac{\ln(3)}{12}$