Need to check simplification of expression with infinite sum of exponentials

Question

In reviewing a paper, I've come across a simplification the looks fishy to me, but I'm having a hard time checking it. I pulled out my old CRC handbook, but neither that nor Google are proving to be very helpful. The author writes the following:

$ \begin{align} T &= \sum_{n=0}^\infty (n + 1) T_1 [e^{-n \lambda_1 T_1} - e^{-(n+1) \lambda_1 T_1}] \\ &= \sum_{n=0}^\infty T_1 e^{-n \lambda_1 T_1} ~~~~~~ (really?) \\ &= \frac{T_1}{1 - e^{-\lambda_1 T_1}} \end{align} $

Can anybody help confirm that the simplification above really is valid?

Answer 1

I agree with the result, and you can see it immediately, without doing any summation, by comparing

$$\sum_{n=0}^{\infty} (n+1) e^{-(n+1) \lambda_1 T_1} $$

and

$$\sum_{n=0}^{\infty} n e^{-n \lambda_1 T_1} $$

These are the same thing; just one is index shifted from the other by $1$. Subtracting these leaves you with the result.

Answer 2

The first step is due to a telescoping sum(well somewhat). The second is obviously a GP. I am assuming $T_1 > 0$ Write out the first k terms and look at the pattern:

$$ T_1\sum_{n=0}^{k-1}(n+1)[e^{-n\lambda_1T_1} - e^{-(n+1)\lambda_1T_1}]$$ $$ = T_1[ 1 - e^{-\lambda_1T_1} + 2e^{-\lambda_1T_1} -2e^{-2\lambda_1T_1}+3e^{-\lambda_1T_1}... +ke^{-(k-1)\lambda_1T_1} -ke^{-k\lambda_1T_1}$$ $$= T_1[ 1 +(-e^{-\lambda_1T_1} + 2e^{-\lambda_1T_1}) +(-2e^{-2\lambda_1T_1}+3e^{-\lambda_1T_1})... +(-(k-1)e^{-(k-1)\lambda_1T_1}+ke^{-(k-1)\lambda_1T_1}) -ke^{-k\lambda_1T_1}$$ $$= T_1 [\sum_{n=0}^{k-1}e^{-n\lambda_1T_1}] - T_1ke^{-k\lambda_1T_1}$$

Now take limits on both sides, noting that $\lim_{k \rightarrow\infty} ke^{-k\lambda_1T_1}=0$, you get the result u wanted. QED

Answer 3

$$T=\sum_{n=0}^\infty (n+1)T_1 [e^{-n\lambda_1T_1} - e^{-(n+1)\lambda_1T_1}]$$ $$=\sum_{n=0}^\infty (n+1)T_1 e^{-n\lambda_1T_1} - \sum_{n=0}^\infty (n+1)T_1 e^{-(n+1)\lambda_1T_1}$$ $$=\sum_{n=0}^\infty (n+1)T_1 e^{-n\lambda_1T_1} - \sum_{n=1}^\infty nT_1 e^{-n\lambda_1T_1}$$ $$=T_1 + \sum_{n=1}^\infty T_1 e^{-n\lambda_1T_1}$$ $$=\sum_{n=0}^\infty T_1 e^{-n\lambda_1T_1}$$