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?
 A: Hint. There is no closed formula here. Compute the first few terms and compare them with the $n$th-harmonic number $H_n=\sum_{k=1}^n\frac{1}{k}$. What can we conjecture? 
P.S. BTW the linked sum $\sum(-1)^k{n\choose k}\frac{1}{k+1}$ is "similar" but quite much easier (it has a closed formula).
A: 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 .
A: This problem and its type appear  at MSE regularly. Suppose we seek to
compute
$$S_n = \sum_{k=1}^n {n\choose k} \frac{(-1)^{k+1}}{k}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}.$$
We then obtain for $1\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^{n+1} \frac{n!}{k^2} \prod_{q=1}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^{n+1} \frac{n!}{k}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^{k+1}}{k}.$$
This means that
$$S_n = \sum_{k=1}^n \mathrm{Res}_{z=k} f(z)$$
and since residues sum to zero we have
$$S_n + \mathrm{Res}_{z=0} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
We can compute  the residue at infinity by inspection  (it is zero) or
more formally through
$$\mathrm{Res}_{z=\infty}
n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^2}
z^2 \prod_{q=1}^n \frac{1}{1/z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0}
\prod_{q=1}^n \frac{z}{1-qz}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} z^n
\prod_{q=1}^n \frac{1}{1-qz} = 0.$$
We get for the residue at $z=0$ that
$$\mathrm{Res}_{z=0} f(z) =
n! (-1)^{n+1}
\left. \left(\prod_{q=1}^n \frac{1}{z-q}\right)'\right|_{z=0}
\\ = - n! (-1)^{n+1} \left.
\left(\prod_{q=1}^n \frac{1}{z-q}\right)
\sum_{q=1}^n \frac{1}{z-q} \right|_{z=0}
\\ = n! (-1)^n \frac{(-1)^{n}}{n!} \left(-H_{n}\right)
= -H_n.$$
We thus have $S_n - H_n = 0$ or
$$\bbox[5px,border:2px solid #00A000]{
S_n = H_n = \sum_{k=1}^n \frac{1}{k}.}$$
