Prove that $\sum^\infty_{n\text{ = }m}\big(\frac{a}{e^{n+b}}\big)=\frac{a\,e^{1-\,b\,-\,m}}{e-1}$

Question

Prove that: $$\sum^\infty_{n\text{ = }m}\bigg(\frac{a}{e^{n+b}}\bigg)=\frac{a\,e^{1-\,b\,-\,m}}{e-1}$$ Where $e$ denotes the base of the exponential function.

I was easily able to prove convergence, however I am struggling to reach this convergence value.

Edit: There is apparently nothing special about $e$. The same result holds for $e$ being any real number.

Answer 1

\begin{align} \sum_{n=m}^\infty \left(\frac{a}{e^{n+b}}\right) &= \frac{a}{e^b}\sum_{n=m}^\infty e^{-n}\\ &=ae^{-b}\left(\sum_{n=0}^\infty e^{-n} -\sum_{n=0}^{m-1} e^{-n}\right)\\ &=ae^{-b}\left(\frac{1}{1-e^{-1}} - \frac{1 - e^{-m}}{1-e^{-1}}\right)\\ &=ae^{-b+1}\left(\frac{1}{e-1}-\frac{1-e^{-m}}{e-1}\right)\\ &=a\cdot\frac{e^{1-m-b}}{e-1} \end{align}

Going from the second to the third line uses the (partial) sum formula of the geometric series, here with common ratio $r = e^{-1} < 1$ (which is necessary for the convergence of the infinite sum). The rest is just using how sums work and algebra.

Answer 2

There aren't too many cases we can get the convergence value of a series; one of the big ones is geometric series.

Here, our series can be written

$$ae^{-b}\sum_{n=m}^\infty \left(\frac1e\right)^n = ae^{-b}\frac{e^{-m}}{1-e^{-1}} = \frac{ae^{1-b-m}}{e-1}$$

Answer 3

By using $$ \sum_{n=0}^\infty r^n=\frac{1}{1-r} $$ one has $$\sum^\infty_{n\text{ = }m}\bigg(\frac{a}{e^{n+b}}\bigg)=ae^{-b-m}\sum_{n=0}^\infty e^{-n}=\frac{ae^{-b-m}}{1-e^{-1}}.$$

Answer 4

$$\sum_{n=m}^\infty \frac{a}{e^{n+b}} = ae^{-b}\sum_{n=m}^\infty e^{-n} = ae^{-b}e^{-m} \sum_{n=0}^\infty e^{-n}$$

You should recognize a convergent geometric series here, for which we know the value of the sum.