TITLE: Evaluate the limit of $(n+1)\int_0^1x^n\ln(1+x)\,dx$ when $n\to\infty$
QUESTION [6 upvotes]: Evaluate the following limit : $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right].$$

We have , $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right]$$
$$=\lim_{n\to \infty}\int_0^1\ln(1+x)\,d(x^{n+1})$$Now put , $x^{n+1}=y$. Then , $$=\lim_{n\to \infty}\int_0^1\ln\left(1+y^{\frac{1}{n+1}}\right)\,dy$$Let , $\displaystyle g_n(y)=\ln\left(1+y^{\frac{1}{n+1}}\right)$.

Edit  :
Then , $\displaystyle g(y)=\lim_ng_n(y)=\ln 2$ in $(0,1]$. 
Now , $\displaystyle \sup_{x\in (0,1]}|g_n(y)-g(y)|=\sup_{x\in (0,1]}\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)=0$. ( As , $y^{\frac{1}{n+1}}$ is monotone increasing function in $(0,1]$ , so $\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)$ is also monotone increasing in $(0,1]$ and so it attains its maximum value at $y=1$ . ) So , $\{g_n(y)\}$ converges uniformly to $\ln 2$.

Then ,we can show that $g_n(y)$ converges uniformly to $\ln 2$ in $(0,1]$ and hence the given limit is $\ln 2$.

Is this correct ? Does there any other technique to evaluate the limit ?

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[10px,#ffd]{\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}x^{n}\ln\pars{1 + x}\,\dd x}} =
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}\exp\pars{n\ln\pars{1 - x}}
\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{\infty}\exp\pars{-nx}\ln\pars{2}\,\dd x}
\qquad\pars{~Laplace'\!s\ Method~}
\\[5mm] = &\
\ln\pars{2}\lim_{n \to \infty}\bracks{\pars{n + 1}{1 \over n}} =
\bbx{\ln\pars{2}}
\end{align}

See Laplace's Method.