SRS - Srinivasa Ramanujan Series

This work is all based on Ramanujan's.

Most people will say it is infinity. But really the answer is $-\frac{1}{12}$ , and not $\infty$ ; it has not only been proven by numberphile but has also shown up in physics as well.

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First, we will prove that $\color{#D61F06}{1-1+1-1+1-1+1-1+1-1+1-...=\frac{1}{2}}$

$1-1=0$ and $1-1+1=1$

We don't know where we will stop in the series, so we take the average of our answer $1\cup0$ and the solution should be $\frac{1}{2}$

$\quad$

Next, $1-2+3-4+5-6+7-8+9-...=?$

$\begin{aligned} \text{Let} \, S & = 1-2+3-4+5-6+7-8+9-\cdots \\ \underline{+S} & = \quad \underline{+1-2+3-4+5-6+7-8+\cdots} \qquad \qquad \text{add S to each side} \\ 2S & = \color{#D61F06}{1-1+1-1+1-1+1-1+1-\cdots} \\ 2S & =\frac{1}{2} \\ \color{#3D99F6}{S} & \color{#3D99F6}{= \frac{1}{4}} \end{aligned}$

$\quad$

Finally, we will look at the serie in this problem: $\large\sum_{n=1}^{\infty} n$

$\begin{aligned} \text{Let} S_1 & =\quad \, 1+2+3+4+5+6+7+8+9+\cdots \\ \underline{-S} & =\underline{-(1-2+3-4+5-6+7-8+9-\cdots)} \qquad \qquad \text{subtract S from each side.} \\ S_1-\color{#3D99F6}{S} & = \quad \qquad 4 \quad\;\; +8 \quad\;\; +12 \quad +16+\cdots\\ & = \, \, 4(1+2+3+4+5+\cdots) \\ S_1-\frac{1}{4} & =4(S_1) \\ -\frac{1}{4} & =3S_1 \\ -\frac{1}{12} & =S_1 \quad \blacksquare \end{aligned}$