Why is $\sum^{\infty}_{j=k} (\frac{e}{k})^{j} = (\frac{e}{k})^{k} \sum_{j=0}^{\infty} (\frac{e}{k})^{j}$ when $k \geq 3$? So I have from some lecture slides that if $k \geq 3$ the following is true $$\sum^{\infty}_{j=k} (\frac{e}{k})^{j} = (\frac{e}{k})^{k} \sum_{j=0}^{\infty} (\frac{e}{k})^{j}$$
I have used wolframalpha to check that it actually is true, at least approximately.
So, the way that I understand it is that $(\frac{e}{k})^{k}$ is suppose to cancel out the first $k$ terms of the RHS sum, e.g. if $k=5$ then the first $5$ summands:
$$
(\frac{e}{5})^{0} + (\frac{e}{5})^{1} + (\frac{e}{5})^{2} + (\frac{e}{5})^{3} + (\frac{e}{5})^{4}
$$
Should be cancelled out by $(\frac{e}{5})^{5}$  e.g.
$$
(\frac{e}{5})^{5} * ((\frac{e}{5})^{0} + (\frac{e}{5})^{1} + (\frac{e}{5})^{2} + (\frac{e}{5})^{3} + (\frac{e}{5})^{4} + \sum^{\infty}_{j=5} (\frac{e}{5})^{j}) = \sum^{\infty}_{j=5} (\frac{e}{5})^{j}
$$
But right now I don't see what rules is being applied for this to be true.
 A: For $k \geq 3$, the value of $x=\frac{e}{k}<1$ because $e=2.71\ldots$ Thus
$$
\sum^{\infty}_{j=k} \left(\frac{e}{k}\right)^{j} = \left(\frac{e}{k}\right)^{k} \sum_{j=0}^{\infty} \left(\frac{e}{k}\right)^{j}
$$
is rewritten as
$$
\sum^{\infty}_{j=k} x^{j} = x^{k} \sum_{j=0}^{\infty} x^{j}
$$
as pointed out by Blitzer.
The sum on right hand side is the geometric series $\sum_{j=0}^{\infty} x^{j}
=1/(1-x)$, convergent because $x<1$. The left hand side is the difference of the same geometric series and its $(k-1)$st partial sum, so we obtain
$$\left(\sum_{j=0}^{\infty} x^{j} - \sum_{j=0}^{k-1} x^{j}\right) =x^k\sum_{j=0}^{\infty} x^{j}.$$
Substituting the sum of the geometric series and its partial sum $\sum_{j=0}^{k-1} x^{j}=\frac{1-x^k}{1-x}$ gives the result.
A: $$\begin{align}
\sum^{\infty}_{j=k} \left(\frac{e}{k}\right)^{j} &= \sum_{j=k}^\infty\left(\frac{e}{k}\right)^k\cdot\left(\frac{e}{k}\right)^{j-k}\\
&=\left(\frac{e}{k}\right)^k\cdot\sum_{j=k}^\infty\left(\frac{e}{k}\right)^{j-k}\\
&=\left(\frac{e}{k}\right)^{k} \sum_{i=0}^{\infty} \left(\frac{e}{k}\right)^i
\end{align}$$
Where:

*

*The first equality follows from the fact that, for all $x$, we have $x^ax^b=x^{a+b}$.

*The second equality follows from the fact that, for convergent series, we have $$\sum \alpha\cdot a_i = \alpha\cdot\sum a_i$$

*The third equality follows because it is a simple relabling, introducing a new index $i=j-k$.

