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

Question

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?

Answer 1

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