# Convergent infinite series

• April 16th 2009, 12:18 PM
colgon
Convergent infinite series
Ok
So I have the problem of:
A definition of a convergent infinite series with the sum S is:
Given any small positive number (ε) it possible to find an integeer N so that |S-Sn|<ε for every n>N

Select some ε's and find the corresponding N's for the following series:
http://photos-c.ak.fbcdn.net/hphotos..._2275134_n.jpg

This is how far I got:

http://photos-e.ak.fbcdn.net/hphotos..._7498630_n.jpg

so http://photos-g.ak.fbcdn.net/hphotos..._5731128_n.jpg

Let ε=1/8

Then |S-Sn|<1/8 for some n>N
so |1-Sn|<1/8
sn will always be positive and less than S ie less than 1 so
1-Sn<1/8
Sn>7/8

Now I am stuck... am I doing this right? If so how do I find N?

Any help appreachated.

colgon
• April 16th 2009, 03:55 PM
boomlegboom
You want the difference between your sum (through some N) and the number the series converges to to be less than epsilon.

Let N=3, then your sum is 1 + (1/2) + (1/4) + (1/8) = 1.875

|1.875-2| = .125 = (1/8) (Noting that the series converges to 2)

So at N=3, the difference = epsilon. Therefore, for n > N, the difference will be less than epsilon=(1/8).

Now just pick some more epsilons and see how many terms it takes to get a difference less than it (your chosen epsilon that is!)

Hope that helps.
• April 16th 2009, 05:27 PM
colgon
thanks
Thank you,
I'll give it a go.

EDIT
ok looking at this I don't understand why the sum of the series converges to 2, I thought it was 1?

Also have you changed mod [sum of series - partial sum] with mod of [partial sum - mod of series]? EDIT ok its a mod it don't matter

Wait I see, you have 1 + (1/2) + (1/4) + (1/8) = 1.875
The sum is from n=1 so I think the first term in the series is 1/2 not 1?
• April 16th 2009, 06:17 PM
colgon
is this it?
ok I think i got it.

Assuming the sum of the series is 1 (as shown in original post, please explain to me if I am mistaken.

Then leading on from where the original post left off at Sn>7/8

We are looking for N to satify Sn>7/8 for some N where n>N

Using Sn=a(1-r^n)/(1-r)

0.875<0.5(1-0.5^n)/(1-0.5)
and
0.875<0.5(1-0.5^n)/0.5
and
0.875<1-0.5^n
and
-0.125<-0.5^n
and
0.125<0.5^n
and
ln0.125<nln0.5
and
(ln0.125)/ln0.5<n
and
3<n

So N=3

EDIT EDIT my inequality manipulation is wrong on this post as I have divided though by -1?

Also can anyone tell me if my mod and inequalitiy manipulation in the original post is correct????

Soz this is a mess... it must be late and time for bed, if you have managed to read though all this an you have any answers please let me know.
• April 16th 2009, 08:28 PM
boomlegboom
My mistake, didn't notice that the sum started at n=1..

Sn is the sum of the series up to the nth term, so

S1 = 1/2
S2 = 1/2 + 1/4 = .75
S3 = 1/2 + 1/4 + 1/8 = .875
...and so on.

The definition of convergence says that for any ε (so someone could pick it as small as they want to) there exists an N (integer: 1,2,3,..) such that for n>N |S-Sn|<ε. So if they said ε = 1, |S-S1|=|1 -1/2|=1/2<1=ε. So N=1, and for any n>1, |S-Sn|<ε is still true.

Now, say ε = 1/100. Your job (to prove convergence) is to find the n such that |1-Sn|<1/100.

Sn > 1-1/100 = 0.99

So, S7 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = 0.9921875 > 0.99

Therefore, N>=7 satisfies |1-Sn|<1/100.

The question really just boils down to how many (N=?) terms of the series are necessary before the sum is within ε of S=1.
• April 16th 2009, 10:32 PM
matheagle
This is a geometric sum, so you can calculate the $n^{th}$ term, EVEN if the infinite series doesn't converge.

Let $S_n=\sum_{k=0}^n a^k$.

Then $S_n=1+\cdots +a^n$ and ${S_n-1\over a}+ a^n=S_n$ when a is not zero.

Thus $S_n(1-a)=1-a^{n+1}$, giving us $S_n={1-a^{n+1}\over 1-a}$.

NOW, if $-1 the infinite series, which is the limit of the above series, converges to ${1\over 1-a}$.
• April 17th 2009, 04:47 AM
colgon
So i think i have it
LEt ε=1/100
So
|S-Sn|<1/100
⇒|1-Sn|<1/100
As 0<Sn<1 execpt at limit we can say
1-Sn<1/100
⇒-Sn<-99/100
⇒Sn>99/100

Now given Sn=a(1-r^n)/(1-r)

⇒0.5(1-0.5^n)/0.5>99/100
⇒1-0.5^n>99/100
⇒-0.5^n>-1/100
⇒0.5^n<1/100
⇒nln0.5<ln1/100
⇒n>(ln1/100)/ln0.5
⇒n>6.64...

If N ∈ ℤ
then to satisfy n>N

N=6

Correct??
• April 17th 2009, 05:34 AM
boomlegboom
Since n>6.64, the next greatest integer is 7, so that would be your N.
• April 17th 2009, 05:36 AM
colgon
n>=7
or
n>6 ?
• April 17th 2009, 08:17 AM
boomlegboom
the definition says for n>=N, so n>=7.