I am stuck on the following problem:
Find the value of : $$\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx].$$
My attempts: Let $$I_{n}= \lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]=\lim_{n\to\infty}[\ln(2)-\int_{0}^{1}\frac{x^{n+1}}{1+x}dx]$$ and now i can not progress. Please help. Thanks in advance for your time
As you've shown, using integration by parts, one has $$\lim_{n\to\infty}\int_0^1x^n\ln(1+x)\:dx=\ln(2)-\lim_{n\to\infty}\int_0^1\frac{x^{n+1}}{1+x}\:dx.$$ Hence, one we calculate the limit of the integral, we'll have the solution. Note that $$\left|\int_0^1\frac{x^{n+1}}{1+x}\:dx\right|\leq\int_0^1\left|x^{n+1}\right|\:dx.$$ We have this bound since $1+x\geq1$, so $\frac{1}{1+x}\leq1$. Now it is easy to see that the integral on the right-hand side has a limit of zero, and therefore $$\lim_{n\to\infty}\int_0^1x^n\ln(1+x)\:dx=\ln(2).$$
Hint: Now you need to prove the second term $\int_0^1 \frac{x^{n+1}}{1+x}dx$ has limit zero when $n$ goes to infinity.
Try to bound it from above with another simpler integral that goes to zero when $n\to \infty$.
You can also use power series. Here's how Euler probably would have done it. I say that because the proof as I present it here is a bit informal but it can be patched up easily. We do term by term integration and then later interchange a limit and an infinite sum. Start with
$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots$
Replace $x$ with $-x$
$\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots$
Integrate term by term (we will use this log series again at the bottom)
$\ln(1+x)=\int_0^x \frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$
Multiply by $x^n$
$x^n\ln(1+x)=x^{n+1}-\frac{x^{n+2}}{2}+\frac{x^{n+3}}{3}-\frac{x^{n+4}}{4}+\cdots$
Integrate it term by term again
$\int_0^1 x^n\ln(1+x)=\frac{1}{n+2}-\frac{1}{2(n+3)}+\frac{1}{3(n+4)}-\frac{1}{4(n+5)}+\cdots$
Multiply by $(n+1)$
$(n+1)\int_0^1 x^n\ln(1+x)=\frac{n+1}{n+2}-\frac{n+1}{2(n+3)}+\frac{n+1}{3(n+4)}-\frac{n+1}{4(n+5)}+\cdots$
Now take the limit as $n$ tends to infinity
$$\lim_{n\rightarrow\infty} (n+1)\int_0^1 x^n\ln(1+x)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\cdots$$
which is the alternating harmonic series. In case you don't remember that this converges to $\ln(2)$ then go back to that natural log series and plug in $x=1$.
Here's another way. Let $M$ be the maximum of $n+1$ independent uniform(0,1) random variables. Then, for $0<x<1$ we have $P(M\leq x)=x^{n+1}$, and differentiating gives the density function of $M$: $f_M(x)=(n+1)x^n$. Also for any $x<1$, we have $P(M\leq x)=x^{n+1}\to 0$, showing that $M\to 1$ in distribution.
Therefore, $$\int_0^1(n+1) x^n \log(1+x)\,dx=\mathbb{E}(\log(1+M))\to\log(2),$$ by the continuity of $\log(1+x)$.
I've used this trick a couple of times before on this site.
Computing $\lim\limits_{n\to+\infty}n\int_{0}^{\pi/2}xf(x)\cos ^n xdx$
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$
$$environment in the title. It really messes up with the front page. – Asaf Karagila Mar 28 '13 at 20:02