Evaluate $\lim_{n\to\infty}\frac{1^{k}+3^{k}+...+(2n-1)^{k}}{n^{k+1}}$ How I find $$\lim_{n\to\infty}\frac{1^{k}+3^{k}+...+(2n-1)^{k}}{n^{k+1}}$$
I don't know how to write this as a sum so that I use a Riemann integral. Or I can write this in 2 sums?
 A: One method is to use the Stolz theorem wich says that the limit of $\frac{a_n}{b_n}$ is the same as $\frac{a_n-a_{n-1}}{b_n-b_{n-1}}$.
Thus your limit becomes
$$\frac{(2n-1)^k}{n^{k+1}-(n-1)^{k+1}}\to \frac{2^k}{k+1}$$
A: $$\lim_{n\to\infty}\frac{1^{k}+2^k+3^{k}+4^k+...+(2n-1)^{k}+(2n)^k}{n^{k+1}}=\lim_{n\to\infty}\frac{1^{k}+3^{k}+...+(2n-1)^{k}}{n^{k+1}}-2^k\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+(n)^{k}}{n^{k+1}}$$
so
$$\lim_{n\to\infty}\frac{1^{k}+3^{k}+...+(2n-1)^{k}}{n^{k+1}}=2^{k+1}\lim_{n\to\infty}\frac{1^{k}+2^k+3^{k}+4^k+...+(2n-1)^{k}+(2n)^k}{(2n)^{k+1}}+2^k\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+(n)^{k}}{n^{k+1}}$$
and use Riemann sum $$\lim_{n\to\infty}\sum_1^{n}f(\frac{i}{n})\frac1n=\int_0^1f(x)dx$$
A: \begin{align*}
  S_n &= \sum_{j=1}^{n} \frac{(2j-1)^{k}}{n^{k+1}} \\
  &= \frac{1}{n} \sum_{j=1}^{n} \frac{(2j-1)^{k}}{n^{k}} \\
  &= \frac{2^{k}}{n} \sum_{j=1}^{n} \left( \frac{2j-1}{2n} \right)^{k}
\end{align*}
which is the mid-point rule of $\displaystyle 2^{k} \int_{0}^{1} x^{k}\, dx$.
Hence
\begin{align*}
  \lim_{n\to \infty} S_{n} &= 2^{k} \int_{0}^{1} x^{k}\, dx \\
  &= 2^{k} \left[ \frac{x^{k+1}}{k+1} \right]_{0}^{1} \\
  &= \frac{2^{k}}{k+1}
\end{align*}
