# Find the value of : $\lim_{n\to\infty}[(n+1)\int_{0}^{1}x^{n}\ln(1+x)dx]$.

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

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Please avoid $$ environment in the title. It really messes up with the front page. – Asaf Karagila Mar 28 '13 at 20:02 ## 4 Answers As you've shown, using integration by parts, one has$$\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).$$- Thanks a lot sir. I have got it. – user52976 Dec 15 '12 at 5:35 @user52976: Glad to help. – Clayton Dec 15 '12 at 5:53 @user52976: Give him +1 for his help. I did. – Babak S. Dec 15 '12 at 6:34 O sure..+1 from me.. – user52976 Dec 15 '12 at 6:36 Hint: Now you need to prove the second term \int_0^1 \frac{x^{n+1}}{1+x}dx has limit zero when n goes to infinity. Try to bound it from above with another simpler integral that goes to zero when n\to \infty. - You can also use power series. Here's how Euler probably would have done it. I say that because the proof as I present it here is a bit informal but it can be patched up easily. We do term by term integration and then later interchange a limit and an infinite sum. Start with \frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots Replace x with -x \frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots Integrate term by term (we will use this log series again at the bottom) \ln(1+x)=\int_0^x \frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots Multiply by x^n x^n\ln(1+x)=x^{n+1}-\frac{x^{n+2}}{2}+\frac{x^{n+3}}{3}-\frac{x^{n+4}}{4}+\cdots Integrate it term by term again \int_0^1 x^n\ln(1+x)=\frac{1}{n+2}-\frac{1}{2(n+3)}+\frac{1}{3(n+4)}-\frac{1}{4(n+5)}+\cdots Multiply by (n+1) (n+1)\int_0^1 x^n\ln(1+x)=\frac{n+1}{n+2}-\frac{n+1}{2(n+3)}+\frac{n+1}{3(n+4)}-\frac{n+1}{4(n+5)}+\cdots Now take the limit as n tends to infinity$$\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$$which is the alternating harmonic series. In case you don't remember that this converges to \ln(2) then go back to that natural log series and plug in x=1. - thanks a lot sir. It is another wonderful approach to the problem. +1 from me. – user52976 Dec 17 '12 at 14:21 Here's another way. Let M be the maximum of n+1 independent uniform(0,1) random variables. Then, for 0<x<1 we have P(M\leq x)=x^{n+1}, and differentiating gives the density function of M: f_M(x)=(n+1)x^n. Also for any x<1, we have P(M\leq x)=x^{n+1}\to 0, showing that M\to 1 in distribution. Therefore,$$\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 $n \to \infty$

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