Consider these two series $$\sum_{n=1}^{\infty}3\left(\frac {1}{2}\right)^n=3$$ $$\sum_{n=0}^{\infty}3\left(\frac {1}{2}\right)^n=6$$
Everybody knows that there should be a difference,
What I know is, if $\left | r \right | < 1$
$$\sum_{n=1}^{\infty}ar^{n-1}=\frac {a}{1-r}$$
I also can change the index $$\sum_{n=0}^{\infty}ar^{n}=\frac {a}{1-r}$$
I want you to dissect first two examples.
The question is,
Why the answer of the first two series are different?
Thank you!
Why $e_1:=(\sum_{n=1}^{\infty}3(\frac {1}{2})^n) \neq e_2:=(\sum_{n=0}^{\infty}3(\frac {1}{2})^n)$?
Because $e_1 = e_2 - 3(\frac{1}{2})^0 = e_2 - 3$.
And read the comments above, which are more useful than my answer.