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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?

Hint. There is no closed formula here. Compute the first few terms and compare them with the $n$th-harmonic number $H_n=sum_{k=1}^nfrac{1}{k}$. What can we conjecture?

P.S. BTW the linked sum $sum(-1)^k{nchoose k}frac{1}{k+1}$ is "similar" but quite much easier (it has a closed formula).

We can write your sum as
$$
eqalign{
& f(n) = sumlimits_{k = 1}^n {left( { - 1} right)^{,k + 1} left( matrix{
n cr
k cr} right){1 over k}} = cr
& = sumlimits_{k = 0}^{n - 1} {left( { - 1} right)^{,k} left( matrix{
n cr
k + 1 cr} right){1 over {k + 1}}} = sumlimits_{k = 0}^infty {left( { - 1} right)^{,k} left( matrix{
n cr
k + 1 cr} right){1 over {k + 1}}} = cr
& = sumlimits_{k = 0}^infty {t_k } cr}
$$

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
$$
eqalign{
& f(n + 1) = sumlimits_{k = 1}^{n + 1} {left( { - 1} right)^{,k + 1} left( matrix{
n + 1 cr
k cr} right){1 over k}} = cr
& = left( {sumlimits_{k = 1}^{n + 1} {left( { - 1} right)^{,k + 1} left( matrix{
n cr
k cr} right){1 over k}} + sumlimits_{k = 1}^{n + 1} {left( { - 1} right)^{,k + 1} left( matrix{
n cr
k - 1 cr} right){1 over k}} } right) = cr
& = left( {sumlimits_{k = 1}^{n + 1} {left( { - 1} right)^{,k + 1} left( matrix{
n cr
k cr} right){1 over k}} + {1 over {n + 1}}sumlimits_{k = 1}^{n + 1} {left( { - 1} right)^{,k + 1} left( matrix{
n + 1 cr
k cr} right)} } right) = cr
& = sumlimits_{k = 1}^n {left( { - 1} right)^{,k + 1} left( matrix{
n cr
k cr} right){1 over k}} - {1 over {n + 1}}left( {0^{,n + 1} - 1} right) = cr
& = f(n) + {1 over {n + 1}} cr}
$$
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
$$ bbox[lightyellow] {
f(n) = sumlimits_{k = 1}^n {left( { - 1} right)^{,k + 1} binom{n}{k}{1 over k}}
= {1 over {n!}}left[ matrix{ n + 1 cr 2 cr} right]
}$$

Also refer to OEIS seq. A000254 .

This problem and its type appear at MSE regularly. Suppose we seek to
compute

$$S_n = sum_{k=1}^n {nchoose k} frac{(-1)^{k+1}}{k}.$$

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 $1le kle n$

$$mathrm{Res}_{z=k} f(z) =
(-1)^{n+1} frac{n!}{k^2} prod_{q=1}^{k-1} frac{1}{k-q}
prod_{q=k+1}^n frac{1}{k-q}
\ = (-1)^{n+1} frac{n!}{k}
frac{1}{k!} frac{(-1)^{n-k}}{(n-k)!}
= {nchoose k} frac{(-1)^{k+1}}{k}.$$

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

$$bbox[5px,border:2px solid #00A000]{
S_n = H_n = sum_{k=1}^n frac{1}{k}.}$$