The Limit of a Riemann Sum

Question

I am watching this video on how to solve a Riemann Sum with limit going to infinity, but at 3:00~3:10 she totally lost me when she scrolled down and said:

$$ \sum_{i=1}^{n}i = \frac{n(n+1)}{2} $$

Could someone explain how she got this?

Answer 1

You got a little mixed up; the equation you should have is just

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}.$$

There are a few ways to prove this. One way is to write:

$$S=1+2+\dots+n \\ S=n+(n-1)+\dots+1 \\ 2S=(n+1)+(n+1)+\dots+(n+1)=n(n+1).$$

Unfortunately that won't help you with $\sum_{i=1}^n i^2$.

Answer 2

Starting with \begin{align} \sum_{k=0}^{n} t^{k} = \frac{1 - t^{n+1}}{1-t} \end{align} then by differentiating both sides it is seen that \begin{align} \sum_{k=0}^{n} k \, t^{k} = \frac{t(1 - (n+1) t^{n} + n t^{n+1}}{(1-t)^{2}}. \end{align} Not let $t=1$ to obtain \begin{align} \sum_{k=0}^{n} k &= \lim_{t \rightarrow 1} \left\{ \frac{t(1 - (n+1) t^{n} + n t^{n+1})}{(1-t)^{2}} \right\} \\ &= \lim_{t \rightarrow 1} \left\{ \frac{1 -(n+1)^{2} t^{n} + n(n+2) t^{n+1} }{ -2(1-t) } \right\} \\ &= \lim_{t \rightarrow 1} \left\{ \frac{ - n(n+1)^{2} t^{n-1} + n(n+1)(n+2) t^{n} }{ 2 } \right\} \\ &= \frac{n(n+1)}{2} \end{align}