Showing that $\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}$ Consider this double sums $(1)$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}\tag1$$
  Where $H_n$ is the n-th harmonic

An attempt:
Rewrite $(1)$ as
$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\left(H_1+{H_2\over 2}+{H_3\over 3}+\cdots+{H_{k+1}\over k+1}\right)\tag2$$
Recall $$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}={1\over n+1}\tag3$$
Not sure how to continue

We seek to show that
$$\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}.$$
This is
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} \frac{k+1}{n+1} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}$$
or
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} 
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^2}.$$
The LHS is
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} \sum_{k=j}^n (-1)^k {n+1\choose k+1}.$$
Writing
$${n+1\choose k+1} = {n+1\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} (1+z)^{n+1}
\; dz$$
we get range control  (vanishes for $k\gt n$) so we  may write for the
inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+1}
\sum_{k\ge j} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\sum_{k\ge 0} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\frac{1}{1+z}
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n}
\; dz
= (-1)^j {n\choose n-j} = (-1)^j {n\choose j}.$$
We thus have to show that
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j {n\choose j}
= \frac{1}{(1+n)^2}$$
or alternatively
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j 
\frac{j+1}{n+1} {n+1\choose j+1}
= \frac{1}{(1+n)^2}$$
which is
$$\sum_{j=0}^n H_{j+1} (-1)^j 
{n+1\choose j+1}
= \frac{1}{1+n}$$
The LHS is
$$\sum_{j=0}^n (-1)^j 
{n+1\choose j+1} \sum_{q=1}^{j+1} \frac{1}{q}
= \sum_{j=0}^n (-1)^j 
{n+1\choose j+1} \sum_{q=0}^{j} \frac{1}{q+1}
\\ = \sum_{q=0}^n \frac{1}{q+1}
\sum_{j=q}^n (-1)^j {n+1\choose j+1}.$$
We re-use the computation from before to get
$$\sum_{q=0}^n \frac{1}{q+1} (-1)^q {n\choose q}
= \sum_{q=0}^n \frac{1}{q+1} (-1)^q \frac{q+1}{n+1} {n+1\choose q+1}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^q {n+1\choose q+1}.$$
We have reduced the claim to
$$\sum_{q=0}^n (-1)^q {n+1\choose q+1} = 1$$
which holds by inspection or by writing 
$$- \sum_{q=1}^{n+1} (-1)^q {n+1\choose q}
= 1 - \sum_{q=0}^{n+1} (-1)^q {n+1\choose q} = 1 - (1-1)^{n+1} = 1.$$
