Evaluate
$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem .
Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\lim\limits_{n\to \infty}\frac{(2n)^p}{(n+1)^{p+1}-n^{p+1}}$ (and now i was thinking to use the binomial theorem )
$\lim\limits_{n\to \infty}\frac{(2n)^p}{n^{p+1}+{p+1 \choose p}n^p+\dots+1-n^{p+1}}$,which will eventually lead to the answer : $\frac{2}{p+1}$ . Is this correct ?
Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$
calculuslimits
Related Solutions
The limit of the ratio in your title is nonzero. Give me a few minutes to type up the following, based on your funtion $g(x) = x^{-1/3}$
if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} \; g(x) \; dx \; < \; \sum_{j=a}^b \; g(j) \; < \; \int_{a-1}^b \; g(x) \; dx $$
Here is a drawing I made, using the letter $f$ rather than $g$
Well, $g$ is integrable at the origin. Let's try $a=1.$ If that is not satisfactory we can just switch to $a=2$ by putting in some extra terms.
$$ \int_1^{n+1} \; x^{-1/3} \; dx \; < \; \sum_{j=a}^b \; j^{-1/3} \; < \; \int_{0}^n \; x^{-1/3} \; dx $$
an antiderivative of $g$ is $G(x) = \frac{3}{2} x^{2/3}$
$$ \frac{3}{2} \left( (n+1)^{2/3} - 1 \right) \; < \; \sum_{j=1}^n \; j^{-1/3} \; < \; \frac{3}{2} n^{2/3} $$
Good enough. Your denominator is simply $ n^{2/3}.$ We see that $$ L = \frac{3}{2} $$
The limit is $+\infty$.
You made an error in the first step. Applying Stolz-Cesaro, you should rather consider:
\begin{eqnarray*} \frac{\frac 1{\sqrt{n+1}}}{\ln{(n+1)}-\ln n} & = & \frac 1{\sqrt{(n+1)}}\cdot \frac 1{\ln \left(1+\frac 1 n\right)}\\ & = & \frac n{\sqrt{(n+1)}}\cdot \underbrace{\frac{\frac 1n}{\ln \left(1+\frac 1 n\right)}}_{\stackrel{n \to \infty}{\longrightarrow}1} \\ & \stackrel{n \to \infty}{\longrightarrow} & +\infty \end{eqnarray*}
Best Answer
Let $ n $ be a positive integer greater than $ 1 \cdot $
Since $ \sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\sum\limits_{k=1}^{2n}{k^{p}}-\sum\limits_{k=1}^{n}{\left(2k\right)^{p}}=\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=n+1}^{2n}{k^{p}}-2^{p}\sum\limits_{k=1}^{n}{k^{p}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(1-2^{p}\right)\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=1}^{n}{\left(n+k\right)^{p}} \cdot $
We have : $ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\left(1-2^{p}\right)\left(\frac{1}{n}\sum\limits_{k=1}^{n}{\left(\frac{k}{n}\right)^{p}}\right)+\frac{1}{n}\sum\limits_{k=1}^{n}{\left(1+\frac{k}{n}\right)^{p}}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\left(1-2^{p}\right)\int\limits_{0}^{1}{x^{p}\,\mathrm{d}x}+\int\limits_{0}^{1}{\left(1+x\right)^{p}\,\mathrm{d}x} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\frac{1-2^{p}}{p+1}+\frac{2^{p+1}-1}{p+1} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}\underset{n\to +\infty}{\longrightarrow}\frac{2^{p}}{p+1} $