Exercise 1.4.21* (Conundrum 3/8)

Question

Show that if $n$ is a positive integer then 
$$\sum_{k=1}^n \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^n \frac{1}{k}.$$

Richard Ganaye · Accepted Answer

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\begin{proof} Consider the function $f$ defined on $\R$ by
$$f(x) = 
\left\{
\begin{array}{ll}
\frac{1 - (1-x)^n}{x}& 	ext{if } x 
e 0,\
n & 	ext{if } x = 0.
\end{array}
ight.
$$
Despite appearances, $f$ is a polynomial function, because, for $x 
e 0$,
\begin{align*}
f(x) &= \frac{1 - (1-x)^n}{1 - (1-x)} = 1 + (1-x) + (1-x)^2 + \cdots + (1-x)^{n-1},
\end{align*}
and $f(0) = n = 1 + (1-0) + (1-0)^2 + \cdots + (1-0)^{n-1}$.

Moreover, if $x 
e 0$,
\begin{align*}
f(x) &= \frac{1}{x} \left(1-  \sum_{k=0}^n (-1)^k \binom{n}{k} x^k ight)\
&= \sum_{k=1}^n (-1)^{k+1} \binom{n}{k} x^{k-1}, \
\end{align*}
and this equality remains true if $x = 0$. Therefore, for all $x \in \R$,
\begin{align*}
f(x) &= 1 + (1-x) + (1-x)^2 + \cdots + (1-x)^{n-1}\
&= \sum_{k=1}^n (-1)^{k+1} \binom{n}{k} x^{k-1}.
\end{align*}

We compute in two ways $\int_0^1 f(x)\,  \D x$.

First
\begin{align*}
\int_0^1 f(x)\,  \D x &= \int_0^1 \left(1 + (1-x) + (1-x)^2 + \cdots + (1-x)^{n-1} ight) \, \D x\
&=\left[ -(1-x) - \frac{(1-x)^2}{2} - \cdots - \frac{(1-x)^{n}}{n}ight]_0^1\
&= 1+ \frac{1}{2}+ \cdots + \frac{1}{n}.
\end{align*}
Next, 
\begin{align*}
\int_0^1 f(x)\,  \D x &= \int_0^1 \sum_{k=1}^n (-1)^{k+1} \binom{n}{k} x^{k-1} \, \D x\
&=\sum_{k=1}^n (-1)^{k+1} \binom{n}{k} \frac{1}{k}.
\end{align*}
This shows
$$\sum_{k=1}^n \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^n \frac{1}{k}.$$
\end{proof}

Exercise 1.4.21* (Conundrum 3/8)

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Exercise 1.4.21* (Conundrum 3/8)

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