Simplifying $\sum_{m=1}^{N+1}e^{-m}$

Question

I am currently trying to simplify $$\sum_{m=1}^{N+1}e^{-m}$$ I set $n = m-1 \implies m = n + 1$ and get $$\sum_{n=0}^{N+1}e^{-(n+1)} = \sum_{n=0}^{N+1}e^{-n-1} = \frac{1}{e}\sum_{n=0}^{N+1}e^{-n}$$ Here my question is if I have to also reduce the sum $N+1 \to N$ when I change the index from $m=1 \to n=0$. This would then give me $$\sum_{m=1}^{N+1}e^{-m} = \frac{1}{e}\sum_{n=0}^{N}e^{-n} = \frac{1}{e} \cdot \frac{e^{-N}(e^{N+1} - 1)}{e-1} = \frac{e^{-N-1}(e^{N+1} - 1)}{e-1} $$

Answer 1

You don't have to make it so complicated. All you need to find the sum of a geometric series are the first term, the number of terms and the common ratio.

Here, the number of terms is $N+1 - 1 + 1 = N+1$. Remember, the rule is always subtract the smallest index from the largest and add $1$.

The first term ($a$) is $e^{-1}$ and the common ratio ($r$) is $e^{-1}$

The formula for the sum is $\displaystyle S(n) = \frac{a(r^n-1)}{r-1}$

The formula gives the sum as $\displaystyle \frac{e^{-1}(e^{-(N+1)} -1)}{e^{-1}-1} = \frac{e^{-(N+1)}-1}{1-e}$

And this is in fact equivalent to your result (which can be further simplified).

Your result: $\displaystyle \frac{e^{-N-1}(e^{N+1} - 1)}{e-1} = \frac{1-e^{-N-1}}{e-1} = \frac{e^{-(N+1)}-1}{1-e}$, the same answer as mine (opening up the brackets, followed by multiplying top and bottom by $-1$ were the steps used).