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Let $$S_n = \int_{0}^{1} \frac{nx^{n-1}}{1+x}dx$$ for $n >0$. Then as $n \to \infty$ , the sequence $(S_n)_{n>0}$ tends to

  1. $0$

  2. $1/2$

  3. $1$

  4. $+\infty$

$$S_n = \int_{0}^{1} \frac{nx^{n-1}}{1+x}dx$$ put $x= \tan^{2}t$

$dx = 2 \tan t \sec^{2} t dt$

so, $$S_n = \int_{0}^{\pi/4} \frac{n \tan^{2n-2}t}{1+\tan^{2} t} 2 \tan t \sec^{2}t dt$$ $\int_{0}^{\pi/4} 2n \tan^{2n-1}t dt$

$2n\int_{0}^{\pi/4} \tan^{2n-1}t dt$

I don't know how to proceed further to find the limit. Is there any other method to find the limit?

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  • 1
    $\begingroup$ If you just want to proceed by elimination, $$\frac{1}{2} \int_0^1 nx^{n-1}dx \leq S_n \leq \int_0^1 nx^{n-1}dx$$ immediately removes 1. and 4. $\endgroup$ – Clement C. Dec 24 '18 at 7:41
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Integrating by parts we get $S_n=x^{n} \frac 1 {1+x} |_0^{1}+\int_0^{1} x^{n} \frac 1 {(1+x)^{2}}$. Second term tends to $0$ so the answer is $\frac 1 2$. [Second term is $\leq \int_0^{1}x^{n}dx =\frac 1 {n+1}$].

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  • $\begingroup$ Thanks! It is really easy than what i was trying to do. $\endgroup$ – Mathsaddict Dec 24 '18 at 6:46
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A third answer, just to see more techniques.

From the change of variables $u=x^n$ (so that $du = n x^{n-1}dx$), we have $$ S_n = \int_0^1 \frac{du}{1+u^{1/n}} $$ and then, by your favorite convergence theorem for integrals (in my case, the Dominated Convergence Theorem), we have $$ \lim_{n\to\infty} S_n = \int_0^1 du\,\lim_{n\to\infty}\frac{1}{1+u^{1/n}} = \int_0^1 du\cdot \frac{1}{2} = \boxed{\frac{1}{2}} $$

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  • $\begingroup$ I have a little confusion about taking the limit under integral sign. Can we always move the limit sign under integration or there is some condition to do this? $\endgroup$ – Mathsaddict Dec 24 '18 at 8:00
  • $\begingroup$ There are conditions. Here, it's legitimate to do it because of the Dominated Convergence Theorem (it's the result I choose to invoke to justify this step here). @Mathsaddict $\endgroup$ – Clement C. Dec 24 '18 at 8:02
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Since you already received a good answer from Kavi Rama Murthy, I add a few thinks for your curiosity.

The antiderivative $$I_n = \int\frac{n\,x^{n-1}}{1+x}\,dx$$ can be computed using special functions that you will learn sooner or later. From it, the integral $$S_n = \int_{0}^{1} \frac{n\,x^{n-1}}{1+x}\,dx=\frac{1}{2} n \left(\psi \left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)\right)$$ where appears the digamma function (which is "similar" to harmonic numbers). Using their asymptotics, we have $$S_n=\frac{1}{2}+\frac{1}{4 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached as well as an approximation.

For example, using $n=10$, the exact result would be $\approx 0.524877$ while the aboce formula gives $\frac{4199}{8000}\approx 0.524875$.

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In context:

$I_n=\displaystyle{\int_{0}^{1}}\dfrac{nx^{n-1}}{1+x}$.

$\displaystyle{\int_{0}^{1}}\dfrac{nx^{n-1}}{1+1}dx \lt I_n\lt$

$\displaystyle{\int_{0}^{1/2}}\dfrac{nx^{n-1}}{1}dx + \displaystyle {\int_{1/2}^{1}}\dfrac{nx^{n-1}}{1+1/2}dx.$

$1/2< I_n <$

$(1/2)^n +(2/3)[1-(1/2)^n]=(1/3)(1/2)^n+2/3.$

Taking the limit $n \rightarrow \infty:$

$1/2 \le \lim_{n \rightarrow \infty}I_n \le 2/3.$

Assuming one of the answers is correct, it is answer 2).

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