Does the Alternating Binomial Sum Formula Hold for All Positive Integers?

[Euge]

Show that for all positive integers $n$, $$\binom{n}{1} - \frac{1}{2}\binom{n}{2} + \cdots + (-1)^{n-1}\frac{1}{n}\binom{n}{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$$

 

[julian]

Spoiler

We have:

\begin{align*}
-\frac{(1-x)^n}{x} = - \binom{n}{0} \frac{1}{x} + \binom{n}{1} - \binom{n}{2} x + \cdots + (-1)^{n-1} \binom{n}{n} x^{n-1}
\end{align*}

So that

\begin{align*}
\int_0^1 \frac{1 - (1-x)^n}{x} dx = \binom{n}{1} - \frac{1}{2} \binom{n}{2} + \cdots + (-1)^{n-1} \frac{1}{n} \binom{n}{n}
\end{align*}

Making the substitution $y=1-x$, we obtain

\begin{align*}
\int_0^1 \frac{1 - (1-x)^n}{x} dx = \int_0^1 \frac{1 - y^n}{1-y} dy = \int_0^1 [1 + y + y^2 + \cdots y^{n-1}] dy = 1 + \frac{1}{2} + \cdots + \frac{1}{n}
\end{align*}