Series problem w/ questions

• Apr 1st 2007, 08:38 PM
rcmango
Series problem w/ questions

heres the worked problem: http://img143.imageshack.us/img143/66/untitledfh4.png

questions:

Is this an infinite series?

Why do we select the first part of the Series as Sn? What is Sn?

..I understand that 1/2n is the starting number and the limit for this series,
and that it is greater than any Sn, however, why do we add + 1/2?

(for example, this part: 1/2n >= Sn + 1/2)

..also, then why does the above equation then change to: 1/2n >= n * 1/2n = 1/2

Is this series divergent because the statement: 0 >= 1/2 is false.
Is that the only reason it is divergent?

I would really appreciate it, if someone could answer these questions the best they can.

Thanks for any help here.
• Apr 5th 2007, 08:38 PM
rcmango
anyone still able to help here please?
• Apr 5th 2007, 11:11 PM
ecMathGeek
Quote:

Originally Posted by rcmango
anyone still able to help here please?

I'm trying to undersand your question, but I'm confused by the notation. It *looks* wrong. This is probably why no one has responded yet.

For example, is Sn supposed to be read as "S sub n"? In other words, was it written as a capital S with a subscript n?

Also, in your work you show a summation:
SUM{i=1 -> 2n} [1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... + 1/(2n)]

Clearly this summation is wrong. The sum is from i = 1 to i = 2n, but you've written the summation as a sum of term of n instead of being in terms of i. I think it's safe to assume you meant to write the summation:
SUM{i=1 -> 2n} [1/i]

Quote:

Is this an infinite series?
No. It's a sum from 1 to 2n.

Quote:

Why do we select the first part of the Series as Sn? What is Sn?
If I've accurately guessed the notation you should be using, S_n, is the short way of saying "the series sum from 1 to n" whereas S_(2n) is "the series sum from 1 to 2n"

Quote:

(for example, this part: 1/2n > Sn + 1/2)
You misread this. It did not say 1/(2n) > Sn + 1/2, it said
S_(2n) = SUM{i=1 -> 2n} [1/i] = 1 + 1/2 + 1/3 + ... + 1/(2n) >= S_n + 1/2
Shortened, this says
S_(2n) > S_n + 1/2

Also, the "..." indicated that the terms go on.
In this case,
1 + 1/2 + 1/3 + ... + 1/(2n) is the same as saying
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + ... + 1/(n - 2) + 1/(n - 1) + 1/n + 1/(n + 1) + 1/(n + 2) + ... + 1/(2n - 2) + 1/(2n - 1) + 1/(2n)
Even here, I had to use the "..." because it would be impossible to write out every term of this series if n is not defined.

Quote:

..I understand that 1/2n is the starting number and the limit for this series,
and that it is greater than any Sn, however, why do we add + 1/2?
1/(2n) is the starting number and the limit for the series? I'm not sure what you mean by this. However, as far as them adding the + 1/2, here's what they were saying:

S_(2n) > S_n + 1/2

In other words, the sum of the series from 1 to 2n adds up to a number larger than the sum of the series from 1 to n plus + 1/2. (Even by adding 1/2 to the sum of the series from 1 to n, the sum from 1 to 2n will still be greater.)

Quote:

..also, then why does the above equation then change to: 1/2n >= n * 1/2n = 1/2
Be very careful with your notation (I think I've made this point quite a few times). In this case, you need to be wary of "order of operations."
n * 1/2n = 1/2n^2
However, the problem you have is not "n * 1/2n," it's "n * 1/(2n)"
n * 1/(2n) = n/(2n) = 1/2 ... the n's reduced.

Besides this, your question is basically, how do we get 1/2? I'll show you.

S_n = 1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n
S_(2n) = 1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n + 1/(n + 1) + 1/(n + 2) + ... + 1/(2n - 1) + 1/(2n)

Notice that S_n and S_(2n) both have the terms:
1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n

We can subtract S_n from S_(2n) and all of the terms I just wrote will disappear, leaving us with:
S_(2n) - S_n = 1/(n + 1) + 1/(n + 2) + ... + 1/(2n - 1) + 1/(2n)

There are two observations I can make about this summation.
1) I know there are "n" terms there. How do I know this? Because S_(2n) has "2n" terms total, S_n has "n" terms total. If we subtract S_(2n) from S_n, we are left with "2n - n" = "n" terms.
2) I know that 1/(2n) is smaller than 1/(2n - 1) and 1/(2n) is smaller than 1/(2n - 2) and 1/(2n) is smaller than 1/(n + 2) and ... etc. 1/(2n) is a smaller number than all of the rest of the terms in that summation above.

Given these two facts, I know that:
S_(2n) - S_n > n*(1/(2n)) AND n*(1/(2n)) = 1/2
Therefore,
S_(2n) - S_n > 1/2

If I add S_n to both sides, I get
S_(2n) > S_n + 1/2 ... look familiar?

Quote:

Is this series divergent because the statement: 0 >= 1/2 is false.
Is that the only reason it is divergent?
Simply put, this series is not divergent because it is not infinite.
• Apr 8th 2007, 09:40 PM
rcmango
yes, you are correct its a subscript. I'll use S_n from now on. my appoligy.

also, this work was provided for me in lecture, by a professor. this is unfortunate, because the work is wrong!?

i'll try to follow and understand your work here.

thanks for all the help so far.

Quote:

If I've accurately guessed the notation you should be using, S_n, is the short way of saying "the series sum from 1 to n" whereas S_(2n) is "the series sum from 1 to 2n"
awesome. good explanation.

Quote:

Besides this, your question is basically, how do we get 1/2? I'll show you.

S_n = 1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n
S_(2n) = 1 + 1/2 + 1/3 + ... + 1/(n - 1) + 1/n + 1/(n + 1) + 1/(n + 2) + ... + 1/(2n - 1) + 1/(2n)
I've seen here and earlier above where you've used ... + 1/(n - 1) + 1/n
and ... + 1/(2n -1) + 1/(2n)

..however, i did not see any of this in the example, why do you start by adding terms in the series, and then change to start subtracting? ex: (2n - 1)

Theres alot going on in this problem, could you tell me in a sentence or two, whats the goal for this problem, to prove what? or maybe what kind of series test it is. I'm missing the logic i think.

I appreciate all the help you've provided so far, I'm struggling with this type of math.