Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Calculate  $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like  $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?

We can write your sum as
$$
\eqalign{
  & f(n) = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n \cr 
  k \cr}  \right){1 \over k}}  =   \cr 
  &  = \sum\limits_{k = 0}^{n - 1} {\left( { - 1} \right)^{\,k} \left( \matrix{
  n \cr 
  k + 1 \cr}  \right){1 \over {k + 1}}}  = \sum\limits_{k = 0}^\infty  {\left( { - 1} \right)^{\,k} \left( \matrix{
  n \cr 
  k + 1 \cr}  \right){1 \over {k + 1}}}  =   \cr 
  &  = \sum\limits_{k = 0}^\infty  {t_k }  \cr} 
$$
and we can express it in terms of a Hypergeometric function, since
$$
\eqalign{
  & t_0  = \left( \matrix{
  n \cr 
  1 \cr}  \right) = n  \cr 
  & {{t_{k + 1} } \over {t_k }} =  - {{n^{\,\underline {\,k + 2\,} } } \over {\left( {k + 2} \right)\left( {k + 2} \right)!}}
 {{\left( {k + 1} \right)\left( {k + 1} \right)!} \over {n^{\,\underline {\,k + 1\,} } }} =   \cr 
  &  =  - {{\left( {n - 1 - k} \right)} \over 1}{{\left( {k + 1} \right)} \over {\left( {k + 2} \right)\left( {k + 2} \right)}}
 = {{\left( {k - n + 1} \right)\left( {k + 1} \right)} \over {\left( {k + 2} \right)\left( {k + 2} \right)}} \cr} 
$$
Then
$$
f(n) = n\;{}_3F_2 \left( {\left. {\matrix{
   { - n + 1,\;1,\;1}  \cr 
   {2,\;2}  \cr 
 } \;} \right|\;1} \right)
$$
Alternatively, we have that
$$
\eqalign{
  & f(n + 1) = \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n + 1 \cr 
  k \cr}  \right){1 \over k}}  =   \cr 
  &  = \left( {\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n \cr 
  k \cr}  \right){1 \over k}}  + \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n \cr 
  k - 1 \cr}  \right){1 \over k}} } \right) =   \cr 
  &  = \left( {\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n \cr 
  k \cr}  \right){1 \over k}}  + {1 \over {n + 1}}\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n + 1 \cr 
  k \cr}  \right)} } \right) =   \cr 
  &  = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
  n \cr 
  k \cr}  \right){1 \over k}}  - {1 \over {n + 1}}\left( {0^{\,n + 1}  - 1} \right) =   \cr 
  &  = f(n) + {1 \over {n + 1}} \cr} 
$$
i.e.:
$$
\left\{ \matrix{
  f(0) = 0 \hfill \cr 
  f(1) = 1 \hfill \cr 
  f(n + 1) - f(n) = \Delta f(n) = {1 \over {n + 1}} \hfill \cr}  \right.
$$
or
$$
\left\{ \matrix{
  g(n) = n!f(n) \hfill \cr 
  g(0) = 0 \hfill \cr 
  g(1) = 1 \hfill \cr 
  g(n + 1) = \left( {n + 1} \right)f(n) + n! \hfill \cr}  \right.
$$
and this is the recurrence satified by 
$$g(n)=\left[ \matrix{  n+1  \cr   2 \cr}  \right]$$
where $\left[ \matrix{  n  \cr   m \cr}  \right]$ represents the (unsigned) Stirling number of 1st kind.
Thus
$$ \bbox[lightyellow] {  
f(n) = \sum\limits_{k = 1}^n {\left( { - 1} \right)^{\,k + 1} \binom{n}{k}{1 \over k}}
  = {1 \over {n!}}\left[ \matrix{  n + 1 \cr   2 \cr}  \right]
}$$
Also refer to OEIS seq. A000254 .
