Proving a summation inequality up to a value

Question

I have this summation inequality and am not sure how to reduce it: $$\sum_{s=1}^{r+1} \frac{s}{N} \gt \sum_{s=r+1}^{N} \frac{s}{N}$$

We have that $r,s$ range from $1.....N$, where does this inequality change (to $\lt$) as $N \to \infty$? I know the value of $r$ should be $\frac{N}{\sqrt{2}}$ but I am not sure how to show this?

I first tried to use summation expansions but this gave me no luck:

$$\frac{r(r-1)}{2}\gt\frac{N(N+1)}{2}$$ $$N^2+N \lt r^2-r$$

However this leads me no closer and no root two in sight? Could anyone shed any light on this?

Thank you!

Answer 1

We have that in the limit

$$\int_0^t x dx=\int_t^1 x dx \implies \frac12 t^2=\frac12 -\frac12 t^2\implies t=\frac{\sqrt 2}2$$

Answer 2

You can mutiply everything by $N$, i.e., $$\sum_{s=1}^{r+1}s > \sum_{s=r+1}^Ns = \sum_{s=1}^Ns - \sum_{s=1}^rs.$$ This means $$\frac{(r+1)(r+2)}{2} > \frac{N(N+1)}{2}-\frac{r(r+1)}{2}.$$ Multiplying by $2$ and rearranging gives $r^2+2r+1 > \frac{N(N+1)}{2}$, hence $$ r > \frac{\sqrt{N(N+1)}}{\sqrt 2}-1. $$