# Limit of $S_n$ as $n \to \infty$

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?

• 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. – Clement C. Dec 24 '18 at 7:41

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 $$\leq \int_0^{1}x^{n}dx =\frac 1 {n+1}$$].

• Thanks! It is really easy than what i was trying to do. – Mathsaddict Dec 24 '18 at 6:46

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 $$\lim_{n\to\infty} S_n = \int_0^1 du\,\lim_{n\to\infty}\frac{1}{1+u^{1/n}} = \int_0^1 du\cdot \frac{1}{2} = \boxed{\frac{1}{2}}$$

• 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? – Mathsaddict Dec 24 '18 at 8:00
• 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 – Clement C. Dec 24 '18 at 8:02

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 $$\frac{4199}{8000}\approx 0.524875$$.

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 $$n \rightarrow \infty:$$

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