# Calculating the limit $\lim_{n\to\infty}[\sum_{k=0}^{n-1}(-1)^k\frac{1}{n-k}]$

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?

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

• 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? Jan 25 '19 at 15:27
• That's right. You got it. Jan 25 '19 at 15:34

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 :

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

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 $$n\to \infty$$.

• 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! ;-) Jan 30 '19 at 5:04