Evaluating the expression: $\sum\limits_1^n(-1)^{k-1}\frac{n \choose k}{k^2}$ Per the title, I want to evaluate the expression:
$$S = \sum\limits_{k=1}^n(-1)^{k-1}\frac{n \choose k}{k^2}$$
Looked on Approach0 but no luck.
I think it has a nice closed form:
$$S = n^2\sum\frac{1}{i^2}+\left(n\sum \frac{1}{i}\right)^2$$

My attempt:
Using the Binomial theorem:
$$\frac{1-(1-x)^n}{x} = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}{x^{k-1}}$$
Integrate both sides from $0$ to $x$.
$$\int\limits_0^x \frac{1-(1-x)^n}{x}dx = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
For the LHS, let $1-x=u$
EDIT: as pointed out by FDP, this is where the issue was. Limits of the integral need to be changed as well. See answer below for correct version.
$$\int\limits_x^0 \frac{1-(u)^n}{1-u}(-du) = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
$$=>\int\limits_0^x \left(\sum\limits_{k=0}^{n-1}u^{k} \right)du = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
$$=>\sum\limits_{k=1}^{n}\frac{x^k} {k} = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
$$=>\sum\limits_{k=1}^{n}\frac{x^{k-1}} {k} = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k-1}}{k}$$
Integrating both sides from $0$ to $1$,
$$=>\int\limits_0^1\sum\limits_{k=1}^{n}\frac{x^{k-1}} {k} = \int\limits_0^1\sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k-1}}{k}$$
$$=>\sum\limits_{k=1}^{n}\frac{1} {k^2} = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{1}{k^2}\tag{1}$$
Equation (1) is incorrect as evidenced by substituting $n=2$. Where did I go wrong?

Why do I care about this? It comes up in calculating the variance of the generalized coupon collector's problem. See here.
 A: The general expression given in comments can write
$$S_n=\sum\limits_{k=1}^n(-1)^k\frac{n \choose k}{k^2}=\frac{\psi ^{(1)}(n+1)}{2}-\frac{\left(H_n\right){}^2}{2}-\frac{\pi ^2}{12}$$
For large enough values of $n$, you could use asymptotics and get
$$S_n=\frac{1}{12} \left(6 \log ^2\left({n}\right)-12 \gamma  \log
   \left({n}\right)-\pi ^2-6 \gamma ^2\right)-\frac{\log
   \left({n}\right)+\gamma -1}{2 n}+\frac{2 \log
   \left({n}\right)+2 \gamma -9}{24
   n^2}+O\left(\frac{1}{n^3}\right)$$ which is in relative error lower  than $0.1$% if $n \geq 4$ and lower  than $0.01$% if $n \geq 7$ .
A: Per comment by @FDP, I managed to correct the Math. Starting over:
Using the Binomial theorem:
$$\frac{1-(1-t)^n}{t} = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}{t^{k-1}}$$
Integrate both sides from $0$ to $x$.
$$\int\limits_0^x \frac{1-(1-t)^n}{t}dx = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
For the LHS, let $1-t=u$
$$\int\limits_1^{1-x} \frac{1-(u)^n}{1-u}(-du) = \sum\limits_{k=1}^n (-1)^{k-1}{n \choose k}\frac{x^{k}}{k}$$
$$\frac{\sum\limits_{k=1}^n\frac{1-(1-x)^k}{k}}{x} = \sum\limits_{k=1}^n (-1)^{k-1} \frac{n\choose k}{k}x^{k-1}$$
Integrate both sides from $0$ to $1$, we get:
$$\sum\limits_{k=1}^n \frac 1 k \int\limits_0^1 \frac{1-(1-x)^k}{x} dx = \sum \frac{n \choose k}{k^2} (-1)^{k-1}$$
Substituting $1-x=t$ in the integral and expanding the geometric series we get:
$$\sum\limits_{k=1}^n \frac 1 k \sum\limits_{j=1}^k \frac 1 j = \sum \frac{n \choose k}{k^2} (-1)^{k-1} = \sum\limits_{k=1}^n\sum\limits_{j=1}^k \frac {1}{jk}$$

This can very easily be extended to $k^r$ in the denominator:
$$\sum_{k=1}^n(-1)^{k-1}\frac{n\choose k}{k^r}=\sum_{i_1<i_2<\dots <i_r}\frac{1}{i_1 i_2 \dots i_r}$$
and the following code verifies this upto three terms in the denominator:
def binom_trms(n,r):
    summ = 0
    for k in range(1,n+1):
        summ += (-1)**(k-1)*comb(n,k)/k**r
    return summ


def inverses_3(n):
    summ = 0
    for i in range(1,n+1):
        for j in range(1,i+1):
            for k in range(1,j+1):
                summ+=1/i/j/k
    return summ


def inverses_2(n):
    summ = 0
    for i in range(1,n+1):
        for j in range(1,i+1):
            summ+=1/i/j
    return summ

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 1}^{n}\pars{-1}^{k - 1}{{n \choose k} \over k^{2}}} =
\sum_{k = 1}^{n}\pars{-1}^{k - 1}{n \choose k}
\bracks{-\int_{0}^{1}\ln\pars{x}x^{k - 1}\,\dd x}
\\[5mm] = &\
\int_{0}^{1}\ln\pars{x}\sum_{k = 1}^{n}{n \choose k}\pars{-x}^{k}
\,{\dd x \over x} =
\int_{0}^{1}{\ln\pars{x}\bracks{\pars{1 - x}^{n} - 1} \over x}\,\dd x
\\[5mm] & =
\left.\partiald{}{\mu}\int_{0}^{1}\bracks{x^{\mu - 1}\pars{1 - x}^{n} - x^{\mu - 1}} \,\dd x\,\right\vert_{\ \mu\ =\ 0^{\large +}} \\[5mm] = &\
\partiald{}{\mu}\bracks{{\Gamma\pars{\mu}\Gamma\pars{n + 1} \over \Gamma\pars{\mu + n + 1}} - {1 \over \mu}}_{\ \mu\ =\ 0^{\large +}}
\\[5mm] = &\
\partiald{}{\mu}\braces{{1 \over \mu}\bracks{{\Gamma\pars{\mu + 1}
\Gamma\pars{n + 1} \over \Gamma\pars{\mu + n + 1}} - 1}}
_{\ \mu\ =\ 0^{\large +}}
\\[5mm] = &\
{1 \over 2}\,\partiald[2]{}{\mu}\bracks{{\Gamma\pars{\mu + 1}
\Gamma\pars{n + 1} \over \Gamma\pars{\mu + n + 1}} - 1}
_{\ \mu\ =\ 0^{\large +}}
\\[5mm] = &\
\left.{1 \over 2}\,\Gamma\pars{n + 1}\,\partiald[2]{}{\mu}
{\Gamma\pars{\mu + 1}
\over \Gamma\pars{\mu + n + 1}}\right\vert_{\ \mu\ =\ 0^{\large +}} \\[5mm] = &\
\bbx{\large {\pi^{2} \over 12} + {1 \over 2}\,H_{n}^{2} -
{1 \over 2}\,\Psi\, '\pars{n + 1}}
\\[5mm] &\
\end{align}
