Harmonic Sums in disguise

Using the identity $H_n \; = \; \int_0^1 \frac{1-x^n}{1-x}\,dx$ we have $\begin{aligned} \sum_{r=1}^N \binom{N}{r} (-1)^{r-1}H_r & = \int_0^1 \frac{1}{1-x} \sum_{r=1}^N \binom{N}{r} (-1)^{r-1}(1 - x^r)\,dx \\ &= \int_0^1 \frac{1}{1-x}\Big\{ (1-x)^N - 1 - (1-1)^N + 1\Big\}\,dx \; = \; \int_0^1 (1-x)^{N-1}\,dx \; = \; \frac{1}{N} \end{aligned}$ With $N=2017$ , this makes the answer to the question $N = \boxed{2017}$ .

$\displaystyle \text{Proposition \#1:} \large \sum_{m=1}^n {\left(\sum_{k=1}^m{a_k}\right) \binom{n}{m} {(-1)}^{m-1}} = \sum_{m=0}^{n-1}{\binom{n-1}{m} a_{m+1}{(-1)}^m}$

$\text{Proof:}$

LHS can be written as -

$\displaystyle \binom{n}{1}a_1 - \binom{n}{2}(a_1 + a_2) + \binom{n}{3} a_3 - \cdots + {(-1)}^{n-1} \binom{n}{n} (a_1 + a_2 + \cdots + a_n)$

$\displaystyle = \left(\binom{n}{1} - \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n}\right) a_1 +\left(- \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n}\right) a_2 + \cdots + \left( {(-1)}^{n-1} \binom{n}{n}\right) a_n$

Using the fact that $-\binom{n}{0} +\binom{n}{1} - \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n} = 0$ ,

$\displaystyle = \binom{n}{0} a_1 + \left(\binom{n}{0} - \binom{n}{1}\right) a_2 + \cdots + \left(\binom{n}{0} -\binom{n}{1} + \binom{n}{2} - \binom{n}{3} +\cdots +{(-1)}^{n-1}\binom{n}{n-1}\right) a_n$

$\displaystyle \text{Proposition \#2 :} \sum_{r=0}^k {\binom{n}{r} {(-1)}^r} = \binom{n-1}{k} {(-1)}^k$

$\text{Proof:}$

We use induction (since it is very easy to observe using Pascal's triangle addition law).

$\displaystyle S(k=0) : \binom{n}{0} = \binom{n-1}{0}$ . $\text{True}$

$\displaystyle S(k) : \sum_{r=0}^k {\binom{n}{r} {(-1)}^r} = \sum_{r=0}^{k-1} {\binom{n}{r} {(-1)}^r} + \binom{n}{k} {(-1)}^k$

$\displaystyle = \binom{n-1}{k-1} {(-1)}^{k-1} + \binom{n}{k} {(-1)}^k = \binom{n-1}{k} {(-1)}^k$ . $\displaystyle \text{Q.E.D.}\blacksquare$

Coming back to our proof, we use this formula.

$\displaystyle = \binom{n-1}{0} a_1 - \binom{n-1}{1} a_2 + \cdots {-1}^{n-1} \binom{n-1}{n-1} a_n$

$\displaystyle = \sum_{m=0}^{n-1}{\binom{n-1}{m} a_{m+1} {(-1)}^m}$

$\displaystyle \text{Q.E.D.}\blacksquare$

Now if we substitute $a_k = \frac{1}{k}$ ,

$\displaystyle \sum_{m=1}^n {\binom{n}{m} H_m {(-1)}^{m-1}} = \sum_{m=0}^{n-1}{\binom{n-1}{m} \frac{{(-1)}^m}{m+1}} = \frac{1}{n}$