Question: Is there any reason why the following happens to be true: $$\int_{-1}^{0} \frac{N(N+1)}{2} dN = \zeta(-1)$$
Here are some Facts:
- If you compute $\sum_{i=1}^{N}i = \frac{N(N+1)}{2}$
- $\int_{-1}^0 \frac{N(N+1)}{2} dN = -\frac{1}{12}$
- for $s>1, \zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$
If you naively plug in $s=-1$ to the above, this is exactly $ \sum_{n=1}^{\infty} n$
$\zeta(-1)=\frac{-1}{12}$
- The $N$th partial sum of this is exactly $\frac{N(N+1)}{2}$
Note this is not a question asking why $\zeta(-1)=\frac{-1}{12}$.