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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?

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You can use the summation equivalent of integration by parts. –  copper.hat Jan 28 '13 at 17:04

3 Answers 3

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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.

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You have to be a little careful when subtracting series. –  Gautam Shenoy Jan 28 '13 at 17:15
1  
In general, yes. In this case...just look. The first term in the second sum is zero, so start with $n=1$, then shift back to $n=0$ and replace $n \rightarrow n+1$ and you get the first sum. Not hard - this is a well-known series, not necessary to re-invent the wheel. –  Ron Gordon Jan 28 '13 at 17:17
    
Correct. Here it is fine. –  Gautam Shenoy Jan 28 '13 at 17:17
    
@rlgordonma, your explanation made good clear sense. Thank you. –  Neil Steiner Jan 28 '13 at 17:21
    
@NeilSteiner: you're welcome. Good luck with the rest of the paper. –  Ron Gordon Jan 28 '13 at 17:23

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

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Gautam, thank you. I regret that I'm not allowed to pick all three answers. –  Neil Steiner Jan 28 '13 at 17:24
    
No worries. The onus is on you to pick the answer that best explained your predicament. –  Gautam Shenoy Jan 28 '13 at 17:25

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

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The catch with this technique is that before you can write $\sum a_n - b_n$ as $\sum a_n - \sum b_n$, you have to show that at least ONE of the limits is finite. –  Gautam Shenoy Jan 28 '13 at 17:12
    
@Alfonso Fernandez, thank you. I regret that I'm not allowed to pick all three answers. –  Neil Steiner Jan 28 '13 at 17:23

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