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?
 A: $$I_n\leq\int_0^1x^ndx=\frac 1{n+1}$$ so $$\lim_{n\rightarrow +\infty}I_n=0$$
A: 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 : 
$$\lim_{n \to \infty}\int_0^1 \frac{x^n}{1+x} \mathrm{d}x = \int_0^1, \lim_{n \to \infty} \frac{x^n}{1+x} \mathrm{d}x = 0$$
A: 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 $n\to \infty$.
