<s>[INST] Integrating to derive relation between binomial co-efficient 
If $C_o,C_1...$ are the binomial coefficents in the expansion of $(1+x)^n$ and $$\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(r+1)^2}  = k \sum_{r=0}^{n} \frac{1}{r+1}$$
Find a k such that the above equation is satisfied

My attempt:
$$ (1+x)^n = \sum_{k=0}^{k=n} \binom{n}{k} x^k$$
integrate both sides
$$ \frac{ (1+x)^{n+1} }{n+1}  =\sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{k+1} +C$$
$$ x= 0 $$
$$  \implies \frac{-n}{n+1}  = C$$
Divide both sides by 'x' and integrate
$$ \int \frac{ (1+x)^{n+1} }{  (x) n+1}  dx = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2} - \frac{n}{n+1} \ln(x)+ C'$$
Or,
$$ \int \frac{ (1+x)^{n+1} }{  (x) n+1}  dx  - \frac{n}{n+1} \ln(x)+ C' = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}$$
Not so sure what do here, like what to put as bounds. Slightly concerned I may have to evaluate a negative logarithm
 [/INST] In trying to evaluate
$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{(r+1)^2}$$
we introduce
$$f(z) = \frac{(-1)^n \times n!}{(z+1)^2} 
\prod_{q=0}^n \frac{1}{z-q}$$
which has the property that for $0\le r\le n$
$$\mathrm{Res}_{z=r} f(z) =
\frac{(-1)^n \times n!}{(r+1)^2}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = \frac{(-1)^n \times n!}{(r+1)^2}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= (-1)^r {n\choose r} \frac{1}{(r+1)^2}.$$
It follows that
$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$
Now residues sum to zero and the residue at infinity of $f(z)$ is zero
by inspection, therefore
$$S_n = - \mathrm{Res}_{z=-1} f(z) =
(-1)^{n+1} \times n! \times
\left. \left( \prod_{q=0}^n \frac{1}{z-q} \right)' \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
\left. \prod_{q=0}^n \frac{1}{z-q} 
\sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
(-1)^{n+1} \prod_{q=0}^n \frac{1}{q+1}
\times \sum_{q=0}^n \frac{1}{q+1}
\\ = n! \times \frac{1}{(n+1)!}
\times \sum_{q=0}^n \frac{1}{q+1}
= \frac{1}{n+1} \sum_{q=0}^n \frac{1}{q+1} = \frac{1}{n+1} H_{n+1}.$$
We see that the desired factor on the sum is $k=\frac{1}{n+1}.$
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