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So, I have got the series: $$I_n = \int_0^1\frac{x^n}{1+x}dx$$ and my task is to find the limit $$\lim_{n\to\infty}(I_n)$$ I have added 1 and subtracted 1 from the numerator and factorised $x^n - 1$ as $(x+1)(x^{n-1}-x^{n-2}+x^{n-3} - ...+(-1)^n)$. Then i did the reduction, and I'm left with calculating the limit: $$\lim_{n\to\infty}[\frac{1}{n}-\frac{1}{n-1}+\frac{1}{n-2} -...+(-1)^{n-1}\frac{1}{2} + (-1)^n] = \lim_{n\to\infty}[\sum_{k=0}^{n-1}(-1)^k\frac{1}{n-k}]$$ I know I can calculate the fraction as a Riemann Sum using the integral of the function $\frac{1}{1-x}$, however I don't know how to deal with the changing signs. How should i approach this? Is there some sort of rule or common practice to deal with alternating signs?

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$$I_n\leq\int_0^1x^ndx=\frac 1{n+1}$$ so $$\lim_{n\rightarrow +\infty}I_n=0$$

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  • $\begingroup$ Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity? $\endgroup$
    – Constantin
    Jan 25 '19 at 15:27
  • $\begingroup$ That's right. You got it. $\endgroup$ Jan 25 '19 at 15:34
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Note that for all $n$ we have :

$$\forall x\in [0,1], \frac{x^n}{1+x} \leq 1$$

A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :

$$\lim_{n \to \infty}\int_0^1 \frac{x^n}{1+x} \mathrm{d}x = \int_0^1, \lim_{n \to \infty} \frac{x^n}{1+x} \mathrm{d}x = 0$$

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Proceeding as in the OP, note that we have two cases

$$\frac{1+x^n}{1+x}=\begin{cases}\sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n\,\,\text{odd}\\\\ \frac2{1+x}-\sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n\,\,\text{even}\end{cases}$$

Integrating over $[0,1]$, letting $n\to \infty$, and using the Taylor series representation for $\log(1+x)$ for $x=1$ reveals

$$\lim_{n\to\infty}\int_0^1 \frac{1+x^n}{1+x}\,dx=\log(2) $$

Hence, we have

$$\lim_{n\to\infty}\int_0^1 \frac{x^n}{1+x}\,dx=\log(2)-\log(2)=0$$


Rather than proceed as in the OP, we write $\frac1{1+x}=\sum_{m=0}^{\infty}(-x)^m$. Then, we see that

$$\begin{align} \int_0^1 \frac{x^n}{1+x} \,dx&=\sum_{m=0}^{\infty} (-1)^m \int_0^1 x^{n+m}\,dx\\\\ &=\sum_{m=n+1}^{\infty}(-1)^{m-n-1} \frac1{m}\tag1 \end{align}$$

As the series $\sum_{m=1}^{\infty}\frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $n\to \infty$.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-) $\endgroup$
    – Mark Viola
    Jan 30 '19 at 5:04

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