<s>[INST] 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
 [/INST] 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 $x=1$.
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