# Math Help - geometric series

1. ## geometric series

given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.

a) set x=1 in the above series and write out the first 4 terms
b) find the exact sum of the series s(1)
c) fing the exact sum of the series s(x)
d) for what values of x does the series from part (c) converge?

for part A i set the S(1) = the sum from n=1 to infinity of (-1/e)^n, and i think i got the first 4 terms but i still don't know what it converges to.

so on part B i got a sum of s(1) = e/(e+1) is it right?

on part C i got S(x) = e^x/(e^x+1) right?

since i am not sure about my answers i don't know part D, for what values it converges.

thanks
K

2. Hello,

Originally Posted by kithy
given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.

a) set x=1 in the above series and write out the first 4 terms
b) find the exact sum of the series s(1)
c) fing the exact sum of the series s(x)
d) for what values of x does the series from part (c) converge?

for part A i set the S(1) = the sum from n=1 to infinity of (-1/e)^n, and i think i got the first 4 terms but i still don't know what it converges to.
You're not asked that... Your series is right.

so on part B i got a sum of s(1) = e/(e+1) is it right?
Right

on part C i got S(x) = e^x/(e^x+1) right?
Only if $e^{-x}<1$ (else, the series diverges), so it's for x>0 (this is part D).

Actually, the exact value of s(x) is : $\sum_{k=1}^{\infty} (-e^{-x})^n=\lim_{N \to \infty} \frac{1+(\frac{1}{e^x})^N}{1+\frac{1}{e^x}}$ (simplify it the way you want)

since i am not sure about my answers i don't know part D, for what values it converges.
You can use the alternate series test

3. Hi

Originally Posted by kithy
given the geometric series s(x) = the sum from n=1 to infinity of (-e^-x)^n.
so on part B i got a sum of s(1) = e/(e+1) is it right?

on part C i got S(x) = e^x/(e^x+1) right?
I don't agree.

$s(1)=\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^n$ is negative because it's an alternating series and $\left(-\frac{1}{\mathrm{e}}\right)^1=-\frac{1}{\mathrm{e}}<0$ so $s(1)=\frac{\mathrm{e}}{1+\mathrm{e}}$ can't be true.

You may try rewriting the series as :

$s(1)=\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^n=-\frac{1}{\mathrm{e}}\sum_{n\geq 1} \left(-\frac{1}{\mathrm{e}}\right)^{n-1}
=-\frac{1}{\mathrm{e}}\underbrace{\left(1-\frac{1}{\mathrm{e}}+\left(-\frac{1}{\mathrm{e}}\right)^2+\left(-\frac{1}{\mathrm{e}}\right)^3+\ldots\right)}_{\fra c{1}{1-\left(-\frac{1}{\mathrm{e}}\right)}}
$

4. Ow, this is the mistake :

The sum of terms of a geometric series is that thing, multiplied by the first term.
Here, the first term is $-\frac 1e$

Hence, $s(1)=-\frac 1e \cdot \frac{e}{1+e}=-\frac{1}{1+e}$