Summation index change I'm currently looking at my teacher's document about convergence but there is something I just can't figure out on my own for some reason.
Here is what he wrote
$$\sum_{n=N}^\infty (\frac{1}{e})^n =(\frac{1}{e})^N\sum_{n=0}^\infty (\frac{1}{e})^n$$ 
And I'm trying to see if my reasonning holds:
$$\sum_{n=0}^\infty (\frac{1}{e})^N = -[1 + (\frac{1}{e})^1 + (\frac{1}{e})^2 + ... + (\frac{1}{e})^(N-1) ] + \sum_{n=0}^\infty (\frac{1}{e})^n $$ 
hence
$$\sum_{n=N}^\infty (\frac{1}{e})^n = \sum_{n=0}^\infty (\frac{1}{e})^(n + N) $$
(because if we start the summation at $n=0$ then we get $  N, (N+1), (N+2)...$ which are the exponents of $\sum_{n=N}^\infty (\frac{1}{e})^n$ )
Therefore 
$$\sum_{n=N}^\infty (\frac{1}{e})^n = \sum_{n=0}^\infty (\frac{1}{e})^(n + N) = \sum_{n=0}^\infty (\frac{1}{e})^n(\frac{1}{e})^N $$ where $N$ is a constant independant of $n$ which leads to
$$\sum_{n=0}^\infty (\frac{1}{e})^n(\frac{1}{e})^N = (\frac{1}{e})^N\sum_{n=0}^\infty (\frac{1}{e})^n $$
So
$$\sum_{n=N}^\infty (\frac{1}{e})^n =(\frac{1}{e})^N\sum_{n=0}^\infty (\frac{1}{e})^n$$ 
However is there a simpler way to see it. Like more intuitive or more direct.. because this result isn't intuitive to me.
Thank you in advance!
 A: Let $a = 1/e$. We want to see why:

$$
\sum_{n=N}^\infty a^n = a^N \sum_{n=0}^\infty a^n
$$

Starting from the RHS, it's easy to see that:
$$
a^N \sum_{n=0}^\infty a^n = \sum_{n=0}^\infty a^{N+n}
$$
Now let's make a substitution. Let $k = N + n$. Notice that if $n \in \{0, 1, 2, \ldots\}$, then $k \in \{N, N + 1, N + 2, \ldots\}$, so it follows that:
$$
\sum_{n=0}^\infty a^{N+n} = \sum_{k=N}^\infty a^k
$$
Finally, $k$ is just a dummy variable used for indexing, so we can rename it back to $n$:
$$
\sum_{k=N}^\infty a^k = \sum_{n=N}^\infty a^n
$$
A: Let $k = n-N$ then $n=N \iff k=0$ and $n \to \infty \iff k \to \infty$. Substituting $n=k+N$:
$$\sum_{n=N}^\infty \left(\frac{1}{e}\right)^n = \sum_{k=0}^\infty \left(\frac{1}{e}\right)^{k+N} = \sum_{k=0}^\infty \left(\frac{1}{e}\right)^k\left(\frac{1}{e}\right)^N = \left(\frac{1}{e}\right)^N\sum_{k=0}^\infty \left(\frac{1}{e}\right)^k$$
Since the summation index is a dummy variable, we can revert it to $n$:
$$
\left(\frac{1}{e}\right)^N\sum_{k=0}^\infty \left(\frac{1}{e}\right)^k = \left(\frac{1}{e}\right)^N\sum_{n=0}^\infty \left(\frac{1}{e}\right)^n
$$
A: You can  memorize the index change
$$\sum^{n}_{k=p}f(k)=\sum^{n-p}_{k=0}f(k+p) $$

"if you subtract "p" in the index, then sum "p" in the argument of the
  function"

$$\sum^{n}_{k=p}f(k)=f(p)+\cdots+f(n) $$
$$\sum^{n-p}_{k=0}f(k+p)=f(0+p)+\cdots+f(n-p+p) .$$
The two forms are the same.
When $\lim n \to \infty$
$$\sum^{\infty}_{k=p}f(k)=\sum^{\infty}_{k=0}f(k+p) $$.
If you have an exponential $f(k)=a^k$, then
$$\sum^{\infty}_{k=p}a^k=\sum^{\infty}_{k=0}a^{k+p}=a^p\sum^{\infty}_{k=0}a^k $$
