Finding the limit of a sequence of integrals

Question

Let us define a sequence of function as $$f_n(x)=\frac{2nx^{n-1}}{x+1}\;\;\text{for each $x\in [0,1]$ and for all $n\in\mathbb{N}$}$$ What is $\displaystyle \lim_{n\to \infty} \int_0^1 f_ n(x) dx$ ?

How to find the limit? If I can interchange the limit with integral the ans is surely $0$. But can we interchange the limit with integral here. All that I know is Lebesgue's monotone convergence theorem and dominated convergence theorem that allow us to interchange limit and integrals. But this seems not to be useful here. Then how to proceed? Any help would be appreciated. Thanks in advance.

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One way to do it is to show that $f_n(x)$ is bounded for $x \in [0,1[$. Hint: for $|x| < 1, \sum_{n=0}^{\infty} nx^n < \infty$. For $x=1$ you surely see that $\lim_{n\to \infty} |f_n(x)| = \infty$ but that of course is a null set and can be ignored for the integral. — Pazu, Feb 16 '18 at 16:04
This is an example of approximation-to-the-identity. indeed, if $f$ is continuous on $[0,1]$, then you can check that $$\lim_{n\to\infty}\int_{0}^{1} nx^{n-1}f(x) , dx = f(1). $$ — Sangchul Lee, Feb 16 '18 at 16:48

score 5 · Answer 1 · answered Feb 16 '18 at 16:39

I thought it would be instructive to present an approach that relies on integration by parts and first-year calculus tools. To that end, we now proceed.

Let $I_n$ be the sequence given by

$$I_n=\int_0^1 \frac{2nx^{n-1}}{1+x}\,dx\tag 1$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac1{1+x}$ and $v=2x^n$ reveals

$$I_n=1+\int_0^1 \frac{2x^n}{(1+x)^2}\,dx\tag2$$

It is easy to see that the integral on the right-hand side of $(2)$ approaches $0$ as $n\to\infty$ (it is bounded below by $\frac1{2(n+1)}$ and bounded above by $\frac2{n+1}$) from which we conclude that

$$\lim_{n\to\infty}I_n=1$$

Robert Z · Accepted Answer · 2018-02-16T16:11:35.217

4

Hint. Note that for $x\in [0,1]$, $$g_n(x):=nx^{n-1}\leq f_n(x)$$ and therefore $$1=\int_0^1 g_n(x) dx \leq \int_0^1 f_ n(x) dx.$$ As regards the other side, $f_n(x)$ is increasing in $[0,1]$ for $n\ge 3$, and for $a\in (0,1)$, $$f_n(x)\leq h_n(x):=\frac{2na^{n-1}}{a+1}\chi_{[0,a]}(x)+\frac{2nx^{n-1}}{a+1}\chi_{(a,1]}(x).$$ Hence $$\int_0^1 f_ n(x) dx\leq \int_0^1 h_ n(x) dx=\frac{2na^{n}}{a+1}+\frac{2(1-a^n)}{a+1}.$$ Can you take it from here?

edited Feb 16 '18 at 16:11

answered Feb 16 '18 at 15:50

Robert Z

147,345

1

Then the limit is >=1 but I cannot see any upper bound for the function.any hint? – mathemagic Feb 16 '18 at 15:57

score 4 · Answer 3 · answered Feb 16 '18 at 19:41

The $nx^{n-1}$ factor suggests we could try using the $y=x^n$ substitution. This way we get \begin{equation} \int_0^1\frac{2}{x+1}nx^{n-1}dx = \int_0^1\frac{2}{x+1}(x^n)'dx = \int_0^1\frac{2}{\sqrt[n]{y}+1}dy. \end{equation} Now, as $\frac{2}{\sqrt[n]{y}+1}\leq2$, we can actually use Lebesgue's dominated convergence theorem to compute the limit. We know that $\sqrt[n]{y}\to1$ a.e. in $[0,1]$, so \begin{equation} \lim_{n\to\infty}\int_0^1\frac{2}{\sqrt[n]{y}+1}dy=\int_0^1\frac{2}{1+1}dy=\int_0^11\,dy=1. \end{equation}

score 3 · Answer 4 · answered Feb 16 '18 at 16:01

3

The limit is 1. For the opposite side, use $$ \int_{0}^{1}f_{n}(x)dx-1 = \int_{0}^{1}\frac{1-x}{1+x}nx^{n-1}dx \leq \int_{0}^{1} (1-x)nx^{n-1}dx = \frac{1}{n+1}\to 0 $$

answered Feb 16 '18 at 16:01

Seewoo Lee

15,670

Finding the limit of a sequence of integrals

4 Answers4