Wikipedia has a proof of the identity $$ H_{n} =\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k}$$
http://en.wikipedia.org/wiki/Harmonic_number#Calculation
Curiously, there is also the identity
$$ \frac{1}{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} H_{k} $$
Can the second identity be derived from the first one?
Best Answer
This is a consequence of binomial inversion. In general, if $f_{n}$ and $g_{n}$ are sequences such that $$f_{n} = \sum_{k=0}^{n}g_{k}{n \choose k}$$
then
$$g_{n} = \sum_{k=0}^{n}(-1)^{n+k}f_{k}{n\choose k}$$
Here I take sequences to be functions from $\mathbb{N} \rightarrow \mathbb{R}$ where I am using the convention that the natural numbers start at $0$.
The second of your identities can be proved from the first using binomial inversion by taking $$f_n = \left\{ \begin{array}{lr} 0 & : n = 0\\ H_{n} & : n > 0 \end{array} \right. $$ and $$g_n = \left\{ \begin{array}{lr} 0 & : n = 0\\ (-1)^{n-1}\frac{1}{n} & : n > 0 \end{array} \right. $$
Several proofs of binomial inversion are given in the answer I linked to.