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$\lim _{n \rightarrow \infty}\left[(n+1) \int_{0}^{1} x^{n} \ln (1+x) d x\right]$

My work $$ \lim _{n \rightarrow \infty} \int_{0}^{1} x^{n} \ln (1+x) d x $$ can be simplified to: $$ \begin{array}{c} \ln (2)-\lim _{n \rightarrow \infty} \int_{0}^{1} \frac{x^{n+1}}{1+x} d x \\ \because\left|\int_{0}^{1} \frac{x^{n+1}}{1+x} d x\right| \leq \int_{0}^{1}\left|x^{n+1}\right| d x \end{array} $$ Now, here the bound $1+x$ is always greater than equal to 1 $$ \Rightarrow \frac{1}{1+x} \leq 1 $$ $$ \therefore \lim _{n \rightarrow \infty} \int_{0}^{1} x^{n} \ln (1+x) d x=\ln (2) $$

Any shorter approach would be highly appreciated!

NB although I have tried to solve this using Dominated Convergence theorem,but couldn't make it

  • This is the exact same as the question: https://math.stackexchange.com/questions/259079/find-the-value-of-lim-n-to-inftyn1-int-01xn-ln1xdx. – JC12 Jul 12 '20 at 07:50

1 Answers1

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Clearly $$ I_n := (n+1) \int_0^1 x^n\, \log(1+x)\, dx \leq \log(2) \int_0^1(n+1)x^n\,dx = \log(2). $$ On the other hand, for every $\epsilon\in (0,1)$ we have that $$ \begin{align} I_n & \geq (n+1) \int_{1-\epsilon}^1 x^n\, \log(1+x)\, dx \\ & \geq \log(2-\epsilon) \int_{1-\epsilon}^1 (n+1) x^n\, dx = \log(2-\epsilon) \left[1 - (1-\epsilon)^{n+1} \right]. \end{align} $$ Since the r.h.s. tends to $\log(2-\epsilon)$ as $n\to +\infty$, we conclude that $\lim_n I_n = \log(2)$.

Rigel
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