find the value of $\int_0^1(C(-y-1)\cdot\sum_{k=1}^{2016}\frac{1}{y+k})dy$ 
The problem: find the value of $\int_0^1 (C(-y-1)\cdot\sum_{k=1}^{2016}\frac{1}{y+k}) dy$, where $C(\alpha$) is the coefficient of $x^{2016}$ in the Maclaurin series for $(1+x)^\alpha$


What I tried:
First of all, substituting $-y-1$ for $\alpha$ I get $C(-y-1) = \frac{-(y+1)(y+2)...(y+2017)}{2016!}$
Then $C(-1-y)\sum_{k = 1}^{2016}\frac{1}{y+k} = \frac{1}{2016!}\sum_{k=1}^{2016}\frac{\prod_{i=1}^{2016}(y+i)}{y+k}$, excuse my wonky formatting. This doesn't seem pretty at all and I'm kind of stuck here.
Questions:
Is there any way to reduce $\sum_{k=1}^{2016}\frac{1}{y+k}$ to something more managable? Like for example how $\sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. 
And if not, how do I even proceed? Thank you!
 A: Whenever I see a problem with a year number in it, I just replace by $N$, i.e., let $N=2016$.
The coefficient of $x^N$ in $(1+x)^{\alpha}$ is 
$$C(\alpha) = \frac1{N!} \prod_{m=0}^{N-1} (\alpha-m)$$
Thus,
$$C(-y-1) = \frac{(-1)^N}{N!} \prod_{m=1}^N (y+m) = \frac{(-1)^N}{N!} p_N(y)$$
Note that
$$p_N'(y)  = p_N(y) \sum_{k=1}^N \frac1{y+k} $$
Thus the integral is equal to
$$\begin{align}I &= \int_0^1 dy \, C(-y-1)  \sum_{k=1}^N \frac1{y+k} \\ &= \frac{(-1)^N}{N!} \int_0^1 dy \,  p_N(y) \sum_{k=1}^N \frac1{y+k} \\ &= \frac{(-1)^N}{N!} \int_0^1 dy \, p_n'(y) \\ &= \frac{(-1)^N}{N!} [p_n(1)-p_n(0)] \\ &= \frac{(-1)^N}{N!} [(N+1)!-N!]  \end{align} $$
Thus,

$$\int_0^1 dy \, C(-y-1)  \sum_{k=1}^{N} \frac1{y+k} = (-1)^N N$$

and in our special case, the answer is $2016$.
A: I think I maybe solved it! Thanks to achille hui's hint and using integration by parts I get: 
$\int_0^1 (\prod_{k=1}^{2017}(y+k))\cdot(\ln(\prod_{k=1}^{2016}(y+k))')dy = -\frac{1}{2016!}(\prod_{k=1}^{2017}(1+k)\cdot\prod_{k=1}^{2016}(1+k) - \prod_{k=1}^{2017}k\cdot\prod_{k=1}^{2016} - \int_0^1\ln(\prod_{k=1}^{2016}(y+k))\cdot(\prod_{k=1}^{2017}(y+k))'dy)$
Let $I$ equal to the first integral, then using integration by parts again I get: 
$I = -\frac{1}{2016!}(\prod_{k=1}^{2017}(1+k)\cdot\prod_{k=1}^{2016}(1+k) - \prod_{k=1}^{2017}k\cdot\prod_{k=1}^{2016}- \ln(\prod_{k=1}^{2016}(1+k))\prod_{k = 1}^{2017}(1+k) -\ln(\prod_{k=1}^{2016}k)\prod_{k=1}^{2017}k - I)$
Then evaluating all of those I get:
$I = -\frac{2017!(2018!-2016!-2017\ln(2016!)-2018\ln(2017))}{1+2016!}$
I don't know if this is the correct answer, but hooray!
