# Finding $\lim_{n \to \infty} \left( 1 + 2\int_0^1 \frac{x^n}{x+1} dx \right)^n$

I have to evaluate $$\lim_{n \to \infty} \left( 1 + 2\int_0^1 \frac{x^n}{x+1} dx \right)^n.$$

My progress: Since $$x \in (0, 1)$$ we can use the series expansion of $$\frac{1}{1+x} = 1-x+x^2-x^3+...$$

Evaluating that integral in the parantheses (which I shall call $$I_n$$) gives

$$I_n = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{n+k} = (-1)^n(\log{2} - A_n)$$

where $$A_n$$ is the nth partial sum of the alternating harmonic series, $$A_n = \sum_{k=1}^n \frac{(-1)^{k+1}}{k}.$$

Since $$\log{2} - A_n$$ goes to 0, it's enough to compute $$2 \lim_{n \to \infty} n(-1)^n(\log{2} - A_n)$$

This is where I got stuck. Any ideas?

Substituting $$x\to t^{\frac{1}{n}}$$ and using $$\lim_{n\to \infty } \, t^{1/n}=1$$ we have

$$I_{n}=\int_0^1 \frac{x^n}{1+x}\,dx = \frac{1}{n}\int_0^1 \frac{t^{\frac{1}{n}}}{1+t^{\frac{1}{n}}}\,dt\to \frac{1}{n}\int_0^1 \frac{1}{1+1}\,dt=\frac{1}{2n}$$

Hence

$$(1+2 I_n)^n \to (1+ \frac{1}{n})^n\to e$$

The limit is e, Euler's constant.

• The replacement of $t^{1/n}$ with $1$ uses the DCT.
– J.G.
Dec 2, 2019 at 17:24
• @ J.G. thanks for the hint. Dec 3, 2019 at 8:29

$$|B_n|=|\int_0^1\dfrac{x^n}{1+x} \text{d}x| \leq \int_0^1 x^n\text{d}x=\dfrac{1}{n+1}$$

$$E_n=(1+2B_n)^n=\exp(n\log(1+2B_n))=\exp(n\times(2B_n+O(B_n^2))$$

so $$\lim E_n=\exp(\lim 2nB_n)$$

and $$nB_n=n\int_0^1 \dfrac{x^n}{1+x} \text{d}x=n\int_0^1\dfrac{u}{1+u^{1/n}}u^{(1-n)/n}\times \dfrac{1}{n}\text{d}u=\int_0^1\dfrac{u^{1/n}}{1+u^{1/n}}\text{d}u$$

so by dominated convergence theorem $$\lim E_n=\exp(1)$$