Evaluating Infinite Series: 3^(n/2) and (-1)^n | Homework Help

[FaraDazed]

Homework Statement

Evaluate:

A: \sum_{n=1}^{\infty} 3^{\frac{n}{2}}

B: \sum_{n=1}^{99} (-1)^n

Homework Equations

The Attempt at a Solution

Really not sure how to show this in the proper mathmatical context so any help or advice is really much appreciated.

A:
\sum_{n=1}^{\infty} 3^{\frac{n}{2}}=1+\frac{1}{\sqrt{3}}+\frac{1}{3}+0.192+\frac{1}{9}
This looks as though the limit is 2, as n tends to infinity

B:
<br /> \sum_{n=1}^{99} (-1)^n=1-1+1-1+1-1<br />
I noted that when n is even it is +1 and when it is odd it is -1 and since 99 is odd, then the limit is 0 as n tends to 99.

 

[SqueeSpleen]

The first series written as you wrote it diverges.
If I take the first numbers of the serie you explicity wrote we have the following serie
\displaystyle \sum_{n=0}^{\infty} \frac{1}{3^{\frac{n}{2}}}
Do you know geometric series?
If | r | &lt; 1 we have the following equality
\displaystyle \sum_{n=0}^{\infty} r^{n} = \frac{1}{1-r}
The other one is not even infinite (so there's no limit), so we have 45 times (-1) and 44 times 1.
Are you sure you didn't write wrong also this one?

 

[FaraDazed]

The first series written as you wrote it diverges.
If I take the first numbers of the serie you explicity wrote we have the following serie
\displaystyle \sum_{n=0}^{\infty} \frac{1}{3^{\frac{n}{2}}}
Do you know geometric series?
If | r | &lt; 1 we have the following equality
\displaystyle \sum_{n=0}^{\infty} r^{n} = \frac{1}{1-r}
The other one is not even infinite (so there's no limit), so we have 45 times (-1) and 44 times 1.
Are you sure you didn't write wrong also this one?

They are both written correctly. Its just my understanding probably, I seriously am very confused by the whole topic of series, sequences, convergence/divergence and limits. We started the topic last week and had 3 50 min lectures on it and it is the first time I have been exposed to it.

No I have not seen that equality, but as that sum is not infinite it does not apply? So, if the summation is not to infinity, then the whole concept of limits does not apply?

EDIT: Can that equality help me in part A? Even though the the exponent is in the denominator of a fraction, and that its not r^n but r^(n/2)?

 

[Mark44]

Homework Statement

Evaluate:

A: \sum_{n=1}^{\infty} 3^{\frac{n}{2}}

B: \sum_{n=1}^{99} (-1)^n

Homework Equations

The Attempt at a Solution

Really not sure how to show this in the proper mathmatical context so any help or advice is really much appreciated.

A:
\sum_{n=1}^{\infty} 3^{\frac{n}{2}}=1+\frac{1}{\sqrt{3}}+\frac{1}{3}+0.192+\frac{1}{9}
This looks as though the limit is 2, as n tends to infinity

You have not expanded this series correctly. The first term is 3^1/2 = √3, not 1. The next term is 3, not 1/√3. You show the last term as being 1/9. There is no last term.

Your book or notes must have some tests for determining whether a series converges or diverges. What tests do you know?

B:
<br /> \sum_{n=1}^{99} (-1)^n=1-1+1-1+1-1<br />

You show this series as having 6 terms. That's incorrect: there are 99 terms.

I noted that when n is even it is +1 and when it is odd it is -1 and since 99 is odd, then the limit is 0 as n tends to 99.

Limits have nothing to do with this problem. This is a finite series. Finding the sum is a straightforward matter of adding up all the terms.

 

[FaraDazed]

You have not expanded this series correctly. The first term is 3^1/2 = √3, not 1. The next term is 3, not 1/√3. You show the last term as being 1/9. There is no last term.

I just realized I entered the problem on here incorrectly. It should be 3^{-\frac{n}{2}} but still had the first term wrong as I started from 0 not 1 by mistake when doing it. And I forget the ... at the end to denote its infinite.

Your book or notes must have some tests for determining whether a series converges or diverges. What tests do you know?

The only ones I know of is the ratio and comparison test, but as the problem sheet has other questions specifically asking us to use those tests I (maybe incorrectly?) assumed they only had to be used on those ones?

You show this series as having 6 terms. That's incorrect: there are 99 terms.

Sorry I am unsure of the correct convention if one is to only write down a few terms of a finite series, or is there not one and if I am going to write the terms then it doesn't make sense unless all of them are there?

Limits have nothing to do with this problem. This is a finite series. Finding the sum is a straightforward matter of adding up all the terms.

OK thanks, it shows how little I have understood of the topic in this very brief introduction to it. What confuses me is if it is a infinite series but the terms eventually go to 0 (or arbitrarily close to it), how is one supposed to find what the sum adds up to if say there were 10000 odd or more terms before it got to 0.

 

[Office_Shredder]

If it's an infinite series you don't find out what the sum adds up to usually, all you are supposed to do is find out whether it converges. Basically the only time when this isn't true is when you have a geometric series, which (a) is an example of. You must have gone over geometric series or seen a formula like
1 + x + x^2 + x^3 +... =\frac{1}{1-x}
at some point in your class.

For part (b), can you calculate the following:
\sum_{n=1}^{4} (-1)^n
\sum_{n=1}^{5} (-1)^n
\sum_{n=1}^{6} (-1)^n
\sum_{n=1}^{7} (-1)^n

and based on that, guess what
\sum_{n=1}^{N} (-1)^n
is as a function of N.

 

[Mark44]

Your book or notes must have some tests for determining whether a series converges or diverges. What tests do you know?

The only ones I know of is the ratio and comparison test, but as the problem sheet has other questions specifically asking us to use those tests I (maybe incorrectly?) assumed they only had to be used on those ones?

One of the first tests that many textbooks present is the "Nth-term test for divergence." It says that if the limit of the nth term of a series is not zero, then the series diverges. Is that one of the tests you've seen?

Sorry I am unsure of the correct convention if one is to only write down a few terms of a finite series, or is there not one and if I am going to write the terms then it doesn't make sense unless all of them are there?

About the only difference is that after the last term you show in an infinite series, you write ..., to indicate that it continues in the same pattern. For a finite series, the usual practice is to write the last term. It's not necessary to write all of the terms in a finite series if there are more than just a few terms in the sum.

OK thanks, it shows how little I have understood of the topic in this very brief introduction to it. What confuses me is if it is a infinite series but the terms eventually go to 0 (or arbitrarily close to it), how is one supposed to find what the sum adds up to if say there were 10000 odd or more terms before it got to 0.

If the terms in an infinite series approach zero, you can't say anything about whether the series converges or diverges.
For example, the first series below converges and the second series diverges, even though the n-th term in each series approaches zero.
$$ \sum_{n = 1}^{\infty} \frac {1} {n^2}$$
$$ \sum_{n = 1}^{\infty} \frac {1} {n}$$

I mentioned the nth term test earlier in this post. Students often misinterpret it as saying, if the nth term goes to zero, then the series converges. This is not generally true. What is true is that if the nth term doesn't go to zero, then the series diverges.

 

[FaraDazed]

If it's an infinite series you don't find out what the sum adds up to usually, all you are supposed to do is find out whether it converges. Basically the only time when this isn't true is when you have a geometric series, which (a) is an example of. You must have gone over geometric series or seen a formula like
1 + x + x^2 + x^3 +... =\frac{1}{1-x}
at some point in your class.

For part (b), can you calculate the following:
\sum_{n=1}^{4} (-1)^n
\sum_{n=1}^{5} (-1)^n
\sum_{n=1}^{6} (-1)^n
\sum_{n=1}^{7} (-1)^n

and based on that, guess what
\sum_{n=1}^{N} (-1)^n
is as a function of N.

The attached image, is all we have been given on geometric series. From that, I was unware that part a was a geometric series. But I don't get how what is in the attached is going to help me solve part a .

http://cdn.imghack.se/medium/2267f471eed35e83f25369f066b5168f.jpg

For those ones above when the summation is to an even number it comes to 0 and the summations to an odd number comes to -1.

I couldn't even guess what that would be as a function of N.

 

[Mark44]

For those ones above when the summation is to an even number it comes to 0 and the summations to an odd number comes to -1.

I couldn't even guess what that would be as a function of N.

What you said in the first paragraph is, I believe, what they're looking for, but could be stated a little better.