Grandi's Series

After the name of Italian monk, priest, philosopher, mathematician and engineer Luigi Guido Grandi (1671-1742), the infinite series $1 - 1 + 1 - 1 \cdots \infty$, is called the Grandi's Series. It is a divergent series, meaning that if we take the partial sums of the series, the limit over the partial sum doesn't exist. Let us say the partial sum is

$S_1 =1 \\  S_2= 1-1=0 \\  S_3 = 1-1+1=1 \\  S_4=1-1+1-1 =0\\ \vdots\\ S_n=\displaystyle\sum_{i=0}^n (-1)^i = 0 \;(\text {if $n$ is odd})\\ \quad\quad\quad \quad\quad\quad= 1 \;(\text {if $n$ is even})\\ \vdots\\ S_\infty=\displaystyle\sum_{i=0}^\infty (-1)^i$

Clearly, it is very difficult to conclude the value of the series. Grandi, however concluded that the sum of the series is $\frac{1}{2}$.

Consider the series,

$S= 1 -1 + 1 -1 \cdots \infty,$ then

$1-S=1- 1 -1 + 1 -1 \cdots \infty = S  \Rightarrow 2S=1 \Rightarrow S=\frac{1}{2}.$

If we take the partial sum of the series and take the mean over it, then we obtain $\frac{1}{2}$. Such summation of infinite series is called Cesaro sum, after the name of another Italian mathematician Ernesto Cesaro (1859 - 1906).

Take the series $1+2+4+8+\cdots \infty$, What is the sum of the series?
Using the similar approach , Say
$c_1 = 1+2+4+8+\cdots \infty \\ \quad= 1+ 2(1+2+4+\cdots \infty)\\ \quad= 1+ 2c_1\\ \Rightarrow c_1-2c_1=1 \Rightarrow c_1=-1.$

These geometric series are quite trickier and gives the values way beyond expectation. Another such series, which we can obtain in similar way using Grandi's series is
$c_2= 1 -2 +3 -4 +5 -6 + \cdots \infty\\ c_2=\quad \;\;1 -2 +3 -4 +5 -6 + \cdots \infty\\ \underline {\hspace{8cm}} \\ 2c_2=1-1+1-1+1-1+\cdots \infty\\ \Rightarrow 2c_2=\frac{1}{2}\\ \Rightarrow c_2= \frac{1}{4}.$

One of the interesting results which Indian mathematician Srinivasa Ramanujan (1887-1920) found in his earlier works is the sum of the series,
$c_3=1+2+3+4+\cdots \infty =\frac{-1}{12}$

We can express $c_3 - c_2$ as
$c_3=1+2+3+4+5+6+\cdots \infty \\   c_2=1-2+3-4+5-6+\cdots \infty\\ -  \;\;  -\;\; +\;\;- \;\; +\;\; \; -\;\; \; +\;\;\;- \cdots \\ \underline {\hspace{8cm}}\\  c_3-c_2=4+8+12+\cdots \infty\\ \Rightarrow c_3-c_2=4(1+2+3+\cdots \infty)\\ \Rightarrow c_3-c_2=4c_3\\ \Rightarrow-3c_3=c_2\\ \Rightarrow c_3=\frac{-c_2}{3}=\frac{-1}{12},\quad \text{as}\; c_2=\frac{1}{4}$