# How to prove $\lim \limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)$ [closed]

I need help to prove this in real analysis. I think it uses IMVT, but not sure how to do it.

Let $f(x)$ be a real valued continuous function on $[0,1]$. Show that $$\lim \limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)$$

• Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. May 29, 2015 at 9:36
• May 29, 2015 at 9:42
• math.stackexchange.com/q/1026197/321264 Dec 29, 2020 at 6:35

$f(x)$ is continuous on a closed interval, so it is bounded, $-M<f(x)<M$ for some $M$. For any $\epsilon>0$, there is a $\delta>0$ for which $f(1)-\epsilon<f(x)<f(1)+\epsilon$ whenever $x>1-\delta$.
$|(n+1)\int_0^{1-\delta}x^nf(x)dx|\leq(n+1)\int_0^{1-\delta}x^nMdx=M(1-\delta)^{n+1}\to0$
Now show that $(n+1)\int_{1-\delta}^1x^nf(x)dx$ approaches a number within $\epsilon$ of $f(1)$.

Since $f$ is continuous, $f$ is bounded on $[0,1]$. Also $$\forall \epsilon>0, \exists \delta>0, \forall x\in(1-\delta,1],\text{there is } |f(x)-f(1)|<\epsilon$$

\begin{align} \left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|&= \left|\int_{0}^{1-\delta} (n+1)x^nf(x)dx+\int_{1-\delta}^{1} (n+1)x^nf(x)dx-f(1)\right| \\ &=\left|f(t_1) \int_{0}^{1-\delta} (n+1)x^ndx+f(t_2)\int_{1-\delta}^{1} (n+1)x^ndx-f(1)\right| \hspace{5 mm} \\ &\hspace{10 mm}(t_1\in(0,1-\delta),t_2\in(1-\delta,1) \text{ and by IMVT}) \\ &=\left|f(t_1)(1-\delta)^{n+1}+f(t_2)(1-(1-\delta)^{n+1})-f(1)\right| \\ &\leqslant 2M(1-\delta)^{n+1}+|f(t_2)-f(1)| \\&<2M(1-\delta)^{n+1}+\epsilon \end{align} So $$\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|\leqslant \varlimsup\limits_{n\to\infty}2M(1-\delta)^{n+1}+\epsilon=\epsilon$$ Since $\epsilon$ is arbitrary small $$\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 \hspace{5 mm} \text{and}\hspace{5 mm} \varliminf\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0$$

So $$\lim\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0\hspace{5 mm} \text{or} \hspace{5 mm}\lim\limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)$$

Consider a much less rigorous process. For the problem \begin{align} \lim \limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1) \end{align} let $f(x)$ be a fuction that can be expanded in a power series, $f(x) = \sum_{k \geq 0} a_{k} \, x^{k}$, that satisfies a real valued continuous function on [0,1] then \begin{align} L &= \sum_{k=0}^{\infty} a_{k} \, \lim \limits_{n\to\infty}(n+1) \, \int_{0}^{1} x^{n+k} \, dx \\ &= \sum_{k=0}^{\infty} a_{k} \, \lim_{n \to \infty} \left( \frac{n+1}{n+k+1} \right) \\ &= \sum_{k=0}^{\infty} a_{k} \, \lim_{n \to \infty} \left( \frac{1 + \frac{1}{n}}{1 + \frac{k+1}{n}} \right) \\ &= \sum_{k=0}^{\infty} a_{k} = f(1) \end{align}

• $f$ is not assumed to be differentiable. So you can not use Taylor series.
– user239780
Jun 28, 2015 at 3:15