Infinite Series

Discussion in 'Math' started by Lightfire, Dec 29, 2013.

1. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello,

I would just like to ask for an explanation of this statement, if it could be step-by-step, I would greatly appreciate it.

I tried by very best but I just can't grasp it. The statement is in blue. Might as well, check for my explanations at the other statement if my explanations were correct or not, or somewhat in between.

Thank you very much AAC world!

EDITED: Peterson, T.S. Calculus with Analytic Geometry. Ken Incorporated, Quezon City.

• Theorem I.pdf
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Last edited: Dec 29, 2013
2. studiot AAC Fanatic!

Nov 9, 2007
5,003
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Well the bit in blue is quite a mouthful isn't it?

In ordinary words it means that every term in the series adds to the sum to n terms so Sn is always increasing.

I think you got that bit, but the next bit is a posh way of saying that there is more than this because every time we add another term the extra that we add to Sn is less than we added previously.

This must be so since although Sn is growing all the time, it never gets larger than some value, K.
If we could add larger and larger amounts then Sn could grow without limit to infinity.

I think this is the key idea you need.

The greek letter epsilon is used to show a small quantity.
So if we consider a small quantity epsilon, no matter how small, added to Sn, which makes Sn+epsilon a bit larger than Sn
If we go far enough along the series we can find a term that adds a smaller amount than epsilon.
Let us call this term N.
Then since every term after N adds even smaller amounts, every term after N adds a smaller amount than epsilon.

So by taking enough terms we can get arbitrarily close to any value of Sn plus epsilon, but never reach it (except at infinity which we do not define).

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3. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
So am I correct in saying that $\sum_{n=1}^{10}u_n$ is much smaller than $\sum_{n=1}^{9}u_n$?

So would that mean that $lim_{n \to\infty} u_{n}=0$?

But since small quantity is not mathematically or technically defined. It could be 100? (I'm joking). But seriously, what will be the value of that? Or by common sense, of course it should be as low as maybe 1 or 0.1 or 0.01, et cetera.

So for example we chose ε to be 0.015625, that's $u_6$, so it means that terms $u_{10}$ until $u_{\infty}$, adds even smaller than ε.

So for example $lim_{n \to\infty} S_{n}=1$. And As I have said I chose ε to be 0.015625 so $S_{n}$ is now 1+015625?

From the PDF I have attached. We can be certain than as $lim_{n \to\infty} S_{n}=1$ Is this what the statement above tells so?

But we're only going to 1 but not 1.015625, which is $S_{n}+1$. Or have I missed something?

Thanks!

4. studiot AAC Fanatic!

Nov 9, 2007
5,003
522
Not at all.

Un is the value of one term of the series and they are all positive so how can the sum of the first 9 terms be greater than the sum of the first ten terms?

Do you understand the sigma notation?

Ʃn = U1 + U2 + U3 +.....Un

What is true is that

(U1 - U2) > (U2-U3) > (U3 - U4)..........> (U(n-1) - Un)

and that

U1 > U2 > U3 > U4 ...............> Un

This is known as a null sequence.

Make sure you are clear about the difference between a sequence and a series.

A sequence is just a list of numbers occurring in order. So there is a first term a second term a third term and so on. We can add terms if we wish but we do not have to.

A series is what we get if we start adding up the terms of the sequence in order, starting with the first one.

This is another way of saying that the limit as n tends to infinity is zero. That is the value of the nth term becomes closer and closer to zero as n approaches infinity.

Note that the nth term (Un) may approach zero but this does not mean that the sum of n (Sn) terms approaches zero.

In your sequence and the sequence, Un = 1/n every term is smaller than any of those before it.
However it is not essential for this to happen for the sequence to be a null sequence.
We say this is sufficient but not necessary.

Another property that your pdf text is stating is this
because the terms get smaller and smaller, there must be some n such that Un < any number we choose. We call the number we choose ε. So yes, epsilon could be 100 if you like, but we would normally use a smaller value.

For example if Un = 1/n then every term after the tenthousandth (n = 10,000) is less than 0.0001.

This sequence Un = (-1)^n/√n does not possess the property that every term is numerically less than the previous one but we can still find an n = N such that the numerical value of the Nth and greater terms is less than any given number (epsilon).
This sequence also converges and we also call it a null sequence.

The property that every term beyond n = a certain N is numerically less than some given number is the necessary condition for a null (convergent) sequence.

Formally

Un is a null sequence if to every positive number ε, there corresponds an integer N such that

|Un| < ε for all values of n greater than N

Last edited: Dec 29, 2013
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5. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Yes. I am familiar with sigma notation. My sloppiness. What I meant is that u10 is much smaller than u9. As we all know if the series converges, the value of un tends to zero as n approaches infinity.