I'm confused on the partial sums formula
Why is
$$\sum_{i=m+1}^\infty \frac{2}{3^i}=\frac{1}{3^m},$$
if $$\sum\limits_{k = 0}^\infty\frac{2}{3^k} = 3.$$
I'm confused on the partial sums formula
Why is
$$\sum_{i=m+1}^\infty \frac{2}{3^i}=\frac{1}{3^m},$$
if $$\sum\limits_{k = 0}^\infty\frac{2}{3^k} = 3.$$
Observe that: $\displaystyle \sum_{i=m+1}^\infty \dfrac{2}{3^i} = \dfrac{1}{3^{m+1}}\cdot \displaystyle \sum_{i=0}^\infty \dfrac{2}{3^i} = \dfrac{3}{3^{m+1}} = \dfrac{1}{3^m}$
Applying the result of the first formula to the second series (where $m=-1$) gives that the second series should sum to $\frac{1}{3^{-1}}=3$. So the second result is consistent with the first.