# How to determine the limit of this sequence?

I was wondering how to determine the limit of $(n^p - (\frac{n^2}{n+1})^p)_{n\in \mathbb{N}}$ with $p>0$, as $n \to \infty$?

For example, when $p=1$, the sequence is $(\frac{n}{n+1})_{n\in \mathbb{N}}$, so its limit is $1$.

But I am not sure how to decide it when $p \neq 1$. Thanks in advance!

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Have you tried a few cases, for example $p=2,3$. It seems that for $p>1$ the sequence diverges. – Beni Bogosel Mar 27 '12 at 13:08

We can write $$a_n=n^p\left(1-\left(1-\frac 1{n+1}\right)^p\right)=n^p\int_{1-\frac 1{n+1}}^ 1pt^{p-1}dt.$$ If $p>1$, the map $t\mapsto t^{p-1}$ is increasing so $a_n\geq n^pp\left(1-\frac 1{n+1}\right)^{p-1}\frac 1{n+1}$ which converges to $+\infty$ and if $0<p<1$ then $a_n\leq n^pp\left(1-\frac 1{n+1}\right)^{p-1}\frac 1{n+1}$ which converges to $0$.

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+1 Nice answer :) – Beni Bogosel Mar 27 '12 at 13:16
I don't know how you came up with this, but it is a smooth answer! – joshin4colours Mar 27 '12 at 17:26
I noticed that $a_n$ is the difference between two terms which are near $n^p$, and writing this difference as an integral we can control it more easily. – Davide Giraudo Mar 27 '12 at 17:27

Using $\frac{1}{1 - x} = 1 + x + \mathcal{O}(x^2)$ for small $x$, we get

$$n^p - \left(\frac{n^2}{n+1}\right)^p = n^p\left(1 - \left(\frac{1}{1 + \frac{1}{n}}\right)^p\right) = n^p\left(1 - \left(1 - \frac{1}{n} + \mathcal{O}(n^{-2})\right)^p\right) = n^p\left(1 - 1 + \frac{p}{n} + \mathcal{O}(n^{-2})\right) = p n^{p-1} - \mathcal{O}(n^{p-2}) \stackrel{n \to \infty}{\longrightarrow} \begin{cases} \infty, & p > 1 \\ 1, & p = 1 \\ 0. & p < 1\end{cases}$$

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Given the Binomial theorem, we have (see Landau notations)

$$\begin{array}{} (n+1)^p-n^p=\Theta(n^{p-1}) & \implies 1-\left(\frac{n}{n+1}\right)^p=\Theta \left( \frac{1}{n} \right) \\ & \implies n^p - \left(\frac{n^2}{n+1}\right)^p=\Theta(n^{p-1}) \end{array}$$

Thus the limit is $0$ for $p<1$, is $1$ for $p=1$ (already computed), and blows up to $\infty$ for $p>1$.

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By Lagrange's theorem, there is a $\xi_n\in(0,1)$ for each $n\in\mathbb N$ such that $(n+1)^p-n^p=p(n+\xi_n)^{p-1}$, so we have:

$$n^p -\left(\frac{n^2}{n+1}\right)^p = \frac{n^p((n+1)^p-n^p)}{(n+1)^p}=\frac{n^pp(n+\xi_n)^{p-1}}{(n+1)^p}=p\left(\frac{n}{n+1}\right)^p(n+\xi_n)^{p-1}.$$

Now, for any $p\in(0,\infty)$, the expression $\left(\frac{n}{n+1}\right)^p$ will converge to $1$. For $p>1$, the expression $(n+\xi_n)^{p-1}$ will converge to $+\infty$ and for $p\in(0,1)$ will converge to $0$. So your original expression will also converge to $+\infty$ and $0$, respectively, in these two cases.

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Asking Wolfram|Alpha, you'll get a series at $n=\infty$: $$n^p\left( \frac{p}{n}+O(1/n^2) \right),$$ so it goes to $\infty$ for $p>1$ and converges to $n^{p-1}p\to 0$ for $0\le p<1$.

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