# Estimating the sum of a series

• Jul 6th 2008, 09:29 PM
auslmar
Estimating the sum of a series
Hello once again,

I have this problem that wants me to estimate the sum of a given series within 0.0001.

"Consider the series $\displaystyle \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."

I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.

:)

Thanks.
• Jul 6th 2008, 11:52 PM
Moo
Hello !

Quote:

Originally Posted by auslmar
Hello once again,

I have this problem that wants me to estimate the sum of a given series within 0.0001.

"Consider the series $\displaystyle \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."

I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.

:)

Thanks.

The sum is equal to :

$\displaystyle 4 \cdot \sum_{k=1}^{\infty} \frac{(-1)^k}{3^k \cdot k!}=4 \cdot \sum_{k={\color{red}1}}^{\infty} \frac{\left(-\frac 13\right)^k}{k!}$.

Now, I will use a known series :

$\displaystyle e^x=\sum_{k={\color{red}0}}^{\infty} \frac{x^k}{k!}=1+\sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}$

$\displaystyle \implies \sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}=e^x-1$

Here, $\displaystyle x=-\tfrac 13$. Can you finish it ? :)
• Jul 7th 2008, 04:37 AM
CaptainBlack
Quote:

Originally Posted by auslmar
Hello once again,

I have this problem that wants me to estimate the sum of a given series within 0.0001.

"Consider the series $\displaystyle \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}$. Estimate the sum of the series within 0.0001."

I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.

:)

Thanks.

This is an alternating series of decreasing terms, so the error in a partial sum $\displaystyle S_n$ of the first $\displaystyle n$ terms is less than the first neglected term.

So:

$\displaystyle |S-S_n|<\frac{4}{3^{n+1} (n+1)!}$

So we want:

$\displaystyle |S-S_n|<0.0001$

we find the smallest integer $\displaystyle n$ such that:

$\displaystyle \frac{4}{3^{n+1} (n+1)!}\le 0.0001$

which by trial and error we find is $\displaystyle n=?$

RonL
• Jul 7th 2008, 06:44 AM
Soroban
Hello, auslmar!

An elaboration of moo's excellent solution . . .

Quote:

Consider the series: .$\displaystyle \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{3^k\cdot k!}$

Estimate the sum of the series within 0.0001.

We have: .$\displaystyle S \;=\;-\frac{4}{3\cdot1!} + \frac{4}{3^2\cdot2!} - \frac{4}{3^3\cdot3!} + \frac{4}{3^4\cdot4!} - \hdots$

. . . . . . . . $\displaystyle S \;=\;4\underbrace{\left[-\frac{\frac{1}{3}}{1!} + \frac{\frac{1}{3^2}}{2!} - \frac{\frac{1}{3^3}}{3!} + \frac{\frac{1}{3^4}}{4!}\hdots\right]}_{\text{Infinite series of: }e^{\text{-}\frac{1}{3}}-1}$

Therefore: .$\displaystyle S\;=\;4\left(e^{-\frac{1}{3}} - 1\right) \;\approx\;-1.1339$

• Jul 7th 2008, 09:11 AM
auslmar
Sorry for the late response, I had to hit the hay.

Thanks to everyone for the help. I'm not sure I actually understand Moo's or Soroban's examples, but CaptainBlack's makes the most sense to me. If I understand correctly, the goal is to determine the sufficient nth term of the series that would be less than or equal to within 0.0001 of the "actual" sum. Is this notion correct? I applied it to a similar problem with correct results.

Thanks again!

Quote:

Originally Posted by CaptainBlack
This is an alternating series of decreasing terms, so the error in a partial sum $\displaystyle S_n$ of the first $\displaystyle n$ terms is less than the first neglected term.

So:

$\displaystyle |S-S_n|<\frac{4}{3^{n+1} (n+1)!}$

So we want:

$\displaystyle |S-S_n|<0.0001$

we find the smallest integer $\displaystyle n$ such that:

$\displaystyle \frac{4}{3^{n+1} (n+1)!}\le 0.0001$

which by trial and error we find is $\displaystyle n=?$

RonL