If $I_n=\int_0 ^1{\frac{x^{n+1}}{x+3}}dx$, prove $\lim_{n \to \infty} n I_n=\frac{1}{4}$

Notice that

$$I_{n+1}-I_n = \int_0^1 \frac{x^{n+1}(x-1)}{x+3} \text{d}x \leq 0$$

so the sequence $\left(I_n\right)_{n\ge 1}$ is decreasing. Also:

$$3I_n+I_{n+1} = \int_0^1 \frac{3x^{n+1}+x^{n+2}}{x+3}\text{d}x = \int_0^1 x^{n+1} \text{d}x = \frac{1}{n+2}$$

Now since the sequence is decreasing we have $I_{n+1} \leq I_{n}$. Therefore:

$$3I_n+I_{n+1}\leq 3I_n+I_n= 4I_n\ \ \ \ \ \ \ \ \ \ (1)$$

Similarly, because $I_n \leq I_{n-1} \Rightarrow 3I_n \leq 3I_{n-1}$, we have:

$$4I_n = 3I_n + I_n \leq 3I_{n-1}+I_n\ \ \ \ \ \ \ \ \ \ (2)$$

Chaining $(1)$ and $(2)$, we get:

$$3I_{n}+I_{n+1} \leq 4I_n \leq 3I_{n-1}+I_{n}$$

or

$$\frac{1}{n+2} \leq 4I_n \leq \frac{1}{n+1}$$

which implies

$$\frac{n}{4(n+2)} \leq nI_n \leq \frac{n}{4(n+1)}$$

From the squeeze theorem, it follows that:

$$\lim_{n\to \infty} nI_n = \frac{1}{4}$$

Put $t=x^{n}$. We get $nI_n=\int_0^{1} \frac {y^{1/n}} {{y^{1/n}+3}} \to \frac 1 4$ because $\frac {y^{1/n}} {{y^{1/n}+3}} \to \frac 1 4$ and $0 <\frac {y^{1/n}} {{y^{1/n}+3}} <1$. [Use DCT to justify interchange of limit and integral].

You can also calculate the limit directly without any recursions using partial integration as follows:

\begin{eqnarray} n\int_0 ^1{\frac{x^{n+1}}{x+3}}dx & = & \int_0^1 \frac{(n+1)x^n\cdot x - x^{n+1}}{x+3}dx\\ & = & \underbrace{\int_0^1 (n+1)x^n \left(1-\frac 3{x+3}\right)dx}_{J_n} - \underbrace{\int_0^1 \frac{x^{n+1}}{x+3}dx}_{\leq \frac 1{3(n+2)}\stackrel{n\to\infty}{\rightarrow}0}\\ \end{eqnarray}

So, it is enough to evaluate $J_n$ using partial integration: \begin{eqnarray} J_n & = & \underbrace{\left[x^{n+1}\left(1-\frac 3{x+3}\right)\right]_0^1}_{=\frac 14} - \underbrace{\int_0^1x^{n+1}\frac{3}{(x+3)^2}dx}_{\leq \frac 1{3(n+2)}\stackrel{n\to\infty}{\rightarrow}0} \\ & \stackrel{n\to\infty}{\rightarrow} & \frac 14 \end{eqnarray}