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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?

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2 Answers 2

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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.

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    $\begingroup$ The replacement of $t^{1/n}$ with $1$ uses the DCT. $\endgroup$
    – J.G.
    Dec 2, 2019 at 17:24
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    $\begingroup$ @ J.G. thanks for the hint. $\endgroup$ Dec 3, 2019 at 8:29
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$$|B_n|=|\int_0^1\dfrac{x^n}{1+x} \text{d}x| \leq \int_0^1 x^n\text{d}x=\dfrac{1}{n+1}$$

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

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 $\lim E_n=\exp(1)$

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