Proof of a sum with binomial coefficients $\sum_{k=1}^n (-1)^{k+1}{\binom nk}\frac{1}{k} = 1 + \frac{1}{2} + \ldots +\frac{1}{n}$ I need to prove:
$$\sum_{k=1}^n (-1)^{k+1}{n \choose k}\frac{1}{k} = 1 + \frac{1}{2} + \ldots +\frac{1}{n}$$ 
for  $n \in \mathbb N$
I should use mathematical induction. So, I've tried going simply by inductive steps.
It's true for $n=1$.
But when I try to evaluate the left side of the term by plugging in the values from $1$ to, say, $5$, I get that the series is actually this:
$$1-1+1-1+1-1+1-1+\ldots$$
Which should at the end equal $0$? But in that case, the problem doesn't make any sense.
Also, by assuming that the term is true for $n=k, k>1$, and then trying to prove that it's also true for $n=k+1$, I can't seem to get anything concrete.
Please help!
 A: By calculating some integrals, we can prove the equation. Note that in some step, there is a change of variable of the the form $y=1-x$ and then renaming $y$ to $x$.
$$\sum_{k=1}^n(-1)^{k+1}{n \choose k}\frac1k=\\
\sum_{k=1}^n(-1)^{k-1}{n \choose k}\int_0^1x^{k-1}dx=\\
\int_0^1\sum_{k=1}^n{n \choose k}(-x)^{k-1}dx=\\
\int_0^1\frac{(1-x)^n-1}{-x}dx=\\
\int_0^1\frac{x^n-1}{x-1}dx=\\
\int_0^1\sum_{k=1}^nx^{k-1}dx=\\
\sum_{k=1}^n\left[\frac{x^k}k\right]_{x=0}^1=\\
\sum_{k=1}^n\frac1k$$
A: You’ve clearly made some error in reasoning or calculation. For $n=4$, say, it’s
$$\binom41\frac11-\binom42\frac12+\binom43\frac13-\binom44\frac14=4-3+\frac43-\frac14=\frac{25}{12}\;,$$
and 
$$1+\frac12+\frac13+\frac14=\frac{12+6+4+3}{12}=\frac{25}{12}\;,$$
just as it’s supposed to be.
In your induction step you should be trying to prove that
$$\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}k\frac1k=\sum_{k=1}^{n+1}\frac1k\tag{1}$$
from the induction hypothesis that 
$$\sum_{k=1}^n(-1)^{k+1}\binom{n}k\frac1k=\sum_{k=1}^n\frac1k\;.\tag{2}$$
The most natural way to try to do this is to add $\frac1{n+1}$ to both sides of $(2)$ and then try to reduce the lefthand side to the lefthand side of $(1)$, i.e., to try to show that
$$\sum_{k=1}^n(-1)^{k+1}\binom{n}k\frac1k+\frac1{n+1}=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}k\frac1k\;.\tag{3}$$
It would certainly help if the binomial coefficients matched up; this suggests that we should use Pascal’s identity on the righthand side of $(3)$:
$$\begin{align*}
\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}k\frac1k&=\sum_{k=1}^{n+1}(-1)^{k+1}\left(\binom{n}k+\binom{n}{k-1}\right)\frac1k\\
&=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n}k\frac1k+\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n}{k-1}\frac1k\;.
\end{align*}$$
Thus, we’re trying to prove that
$$\sum_{k=1}^n(-1)^{k+1}\binom{n}k\frac1k+\frac1{n+1}=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n}k\frac1k+\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n}{k-1}\frac1k\;.\tag{4}$$
The first sum on the righthand side of $(4)$ might as well just run from $k=1$ to $n$, since $\binom{n}{n+1}=0$ anyway. Thus, we can subtract that sum from both sides to find that $(4)$ is equivalent to
$$\frac1{n+1}=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n}{k-1}\frac1k\;.\tag{5}$$
At this point it’s useful to know that 
$$\binom{n}{k-1}\frac1k=\binom{n+1}k\frac1{n+1}\;;$$
I’ll leave that to you to verify. Then an easy application of the binomial theorem will get you what you want.
