Math Help - Simple problem regarding sequences

1. Simple problem regarding sequences

Hello!

I have a "simple" sequence, but I dont know much about sequences. Is there any easy way of computing what is the nth value of the sequence:

Sum[e^(-k i)]

with i going from 0 to n. k is a constant.

Is there any way of computing this besides doing the actual sum?
Thanks for all the help

2. Originally Posted by Sosi
Hello!

I have a "simple" sequence, but I dont know much about sequences. Is there any easy way of computing what is the nth value of the sequence:

Sum[e^(-k i)]

with i going from 0 to n. k is a constant.

Is there any way of computing this besides doing the actual sum?
Thanks for all the help
Hi Sosi,

you can create a formula as follows

$S=e^0+e^{-k}+e^{-2k}+e^{-3k}+....+e^{-(n-1)k}+e^{-nk}$

$=1+e^{-k}\left(e^{-k}+e^{-2k}+....e^{-(n-2)k}+e^{-(n-1)k}\right)
$

In brackets is the sum itself with the last term removed

$S=1+e^{-k}\left(S-e^{-nk}\right)$

$S=1+Se^{-k}-e^{-(n+1)k}$

$S\left(1-e^{-k}\right)=1-e^{-(n+1)k}$

$S=\frac{1-e^{-(n+1)k}}{1-e^{-k}}$

Alternatively, you could use the formula for summing a geometric series.

$S=\frac{a\left(1-r^n\right)}{1-r}$

where "a" is the first term, "r" is the common ratio and "n" is the number of terms.

$a=e^0=1$

$r=\frac{e^{-2k}}{e^{-k}}=e^{-k}$

There are (n+1) terms as i ranges from 0 to n.

$S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}}$

3. Originally Posted by Archie Meade
Hi Sosi,

you can create a formula as follows

$S=e^0+e^{-k}+e^{-2k}+e^{-3k}+....+e^{-(n-1)k}+e^{-nk}$

$=1+e^{-k}\left(e^{-k}+e^{-2k}+....e^{-(n-2)k}+e^{-(n-1)k}\right)
$

In brackets is the sum itself with the last term removed

$S=1+e^{-k}\left(S-e^{-nk}\right)$

$S=1+Se^{-k}-e^{-(n+1)k}$

$S\left(1-e^{-k}\right)=1-e^{-(n+1)k}$

$S=\frac{1-e^{-(n+1)k}}{1-e^{-k}}$

Alternatively, you could use the formula for summing a geometric series.

$S=\frac{a\left(1-r^n\right)}{1-r}$

where "a" is the first term, "r" is the common ratio and "n" is the number of terms.

$a=e^0=1$

$r=\frac{e^{-2k}}{e^{-k}}=e^{-k}$

There are (n+1) terms as i ranges from 0 to n.

$S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}}$
Wow! Thanks! So $S=\frac{1\left(1-e^{-(n+1)k}\right)}{1-e^{-k}}$ will give me the value that I want. For instance, $S(10)=\frac{1\left(1-e^{-(10+1)k}\right)}{1-e^{-k}}$

correct?

Thanks so much!