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$$ \lim_{n\rightarrow \infty} \int_{\frac 1 n }^1 \frac { 1+nx }{ (1+x)^n } \, dx $$

How can I solve this problem using Bounded convergence theorem?

Gamma Function
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user151525
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    If you're using some sort of web software that told you to write \underset =\lim_{ n\rightarrown\rightarrow \infty }{ lim }, you should sue the creators for malpractice. I changed it to \lim_{n\rightarrow \infty}. – Michael Hardy May 17 '14 at 05:21
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    try $u=nx$ and see what happens – clark May 17 '14 at 05:37

2 Answers2

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Hint

We can first look at the antiderivative which is quite simple since

$$\frac { 1+nx }{ (1+x)^n }=\frac {n}{ (1+x)^{n-1}}-\frac {n-1}{ (1+x)^{n}}$$ So, after simplifications, $$I= \int \frac { 1+nx }{ (1+x)^n } \, dx=-\frac{(x+1)^{1-n} (n x+2)}{n-2}$$ From this, it follows that $$J_n= \int_{\frac 1 n }^1 \frac { 1+nx }{ (1+x)^n } \, dx=\frac{\frac{3 \left(\frac{1}{n}+1\right)^{-n} (n+1)}{n}-2^{1-n} (n+2)}{n-2}$$ I am sure that you can take from here

Claude Leibovici
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Write the integral as $$ \int \frac{1 + nx}{(1+x)^n} I_{[1/n,1]}(x) dx = \int f_n(x) dx $$ where the integrand $f_n$ here satisfies $$ 0 \leq f_n(x) \leq I_{[0,1]}(x) $$ since \begin{align} (1 + x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ &= 1 + nx + \sum_{i=2}^n \binom{n}{i} x^i \\ &\geq 1 + nx \end{align} for all non-negative $x$. The function $I_{[0,1]}(x)$ is Lebesgue integrable and hence the limit can be brought in.

Now you have to determine the limit, and since both the numerator and denominator escape to infinity, you have an indeterminate form and can use L'Hopital's suggestion: $$ \lim_{t \to \infty}\frac{1 + tx}{(1+x)^t} = \lim_{t \to \infty}\frac{x}{\ln(1+x)(1+x)^t} = 0. $$

user139388
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