Need to check simplification of expression with infinite sum of exponentials 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?
 A: 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
A: 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.
A: $$T=\sum_{n=0}^\infty (n+1)T_1 [e^{-n\lambda_1T_1} - e^{-(n+1)\lambda_1T_1}]$$
$$=\sum_{n=0}^\infty (n+1)T_1 e^{-n\lambda_1T_1} - \sum_{n=0}^\infty (n+1)T_1 e^{-(n+1)\lambda_1T_1}$$
$$=\sum_{n=0}^\infty (n+1)T_1 e^{-n\lambda_1T_1} - \sum_{n=1}^\infty nT_1 e^{-n\lambda_1T_1}$$
$$=T_1 + \sum_{n=1}^\infty T_1 e^{-n\lambda_1T_1}$$
$$=\sum_{n=0}^\infty T_1 e^{-n\lambda_1T_1}$$
