Determining whether a sequence of partial sums is convergent or divergent

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November 2nd 2012, 10:01 AM
MathIsOhSoHard

Determining whether a sequence of partial sums is convergent or divergent

Given the following sequence of partial sums:
$S_N=\frac{1}{N}cos(N\pi)$

My attempt:
If I can find the series for this sequence of partial sums, then I can test the series to see if it's convergent or divergent.

$a_N=S_N-S_{N-1}=\sum^N_{n=1}a_n-\sum^{N-1}_{n=1}a_n$

Thus:
$a_N=\frac{1}{N}cos(N\pi)-\frac{1}{N-1}cos((N-1)\pi)$

Since $cos(N\pi)=(-1)^N$ :

$a_N=\frac{1}{N}(-1)^N-\frac{1}{N-1}(-1)^{N-1}$

But this is as far as I got.
I would like to express this with a factor $(-1)^N$ so that I can use the alternating series test, but I'm not sure how exactly.

$a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
Would this be correct?
November 2nd 2012, 10:40 AM
Plato

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by MathIsOhSoHard

Given the following sequence of partial sums:
$S_N=\frac{1}{N}cos(N\pi)$

If the real question is contained in the title of this thread, then you have wasted a lot effort.
Because the partial sums are given as $S_N=\frac{1}{N}cos(N\pi)$ .
You know $(S_N)\to 0$ so the series converges to zero.

If the question asks something else, then the title is misleading.
November 2nd 2012, 11:56 AM
MathIsOhSoHard

Re: Determining whether a sequence of partial sums is convergent or divergent

This is the question in its entirety:

For a series $\sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $S_N=\frac{1}{N}cos(N\pi)$ . Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $a_n$ and express it as an alternating series.

(It's supposed to be an infinity sign above the sum, but for some reason LATEX doesn't want to display it)
November 2nd 2012, 12:31 PM
Plato

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by MathIsOhSoHard

This is the question in its entirety:
For a series $\sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $S_N=\frac{1}{N}cos(N\pi)$ . Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $a_n$ and express it as an alternating series.

Well the only difficult part is "Determine whether the series is absolutely convergent,"
For that reason you may want to find $a_n$ .
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$ .

[tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives $\sum\limits_{n = 0}^\infty {ar^n }$
November 2nd 2012, 02:07 PM
MathIsOhSoHard

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by Plato

Well the only difficult part is "Determine whether the series is absolutely convergent,"
For that reason you may want to find $a_n$ .
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$ .

[tex]\sum\limits_{n = 0}^\infty {ar^n } [/tex] gives $\sum\limits_{n = 0}^\infty {ar^n }$

Are you sure $a_n$ can't be written as $a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
November 2nd 2012, 02:21 PM
Plato

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by MathIsOhSoHard

Are you sure $a_n$ can't be written as $a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$

YES

It is $a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

You made a sign error.
November 2nd 2012, 03:02 PM
MathIsOhSoHard

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by Plato

YES

It is $a_N=(-1)^N\cdot \left ( \frac{1}{N}+\frac{1}{N-1}\right )$

You made a sign error.

How do you rewrite $(-1)^{N-1}$ into $(-1)^N$ ? :)
November 2nd 2012, 03:12 PM
Plato

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by mathisohsohard

how do you rewrite $(-1)^{n-1}$ into $(-1)^n$ ? :)

$(-1)^{n-1}=(-1)^n(-1)^{-1}=-(-1)^n$

You see $(-1)^{-1}=\frac{1}{-1}=-1$
November 2nd 2012, 03:37 PM
MathIsOhSoHard

Re: Determining whether a sequence of partial sums is convergent or divergent

Great! Now I get it :)

So by using the alternating series test, it can be shown that the series is convergent.

And to test for absolute convergence, using the comparison test:
$\left | a_n \right | \le \left | b_n \right |$
$\sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $b_n$ is divergent too, thus it is not absolutely convergent.

So conclusion would be that it's conditionally convergent?
Would the absolute value of the fraction be $\frac{1}{n-1}$ or $\frac{1}{n+1}$ ? I wasn't quite sure about that.
November 2nd 2012, 03:44 PM
Plato

Re: Determining whether a sequence of partial sums is convergent or divergent

Quote:

Originally Posted by MathIsOhSoHard

Great! Now I get it :)

So by using the alternating series test, it can be shown that the series is convergent.

And to test for absolute convergence, using the comparison test:
$\left | a_n \right | \le \left | b_n \right |$
$\sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $b_n$ is divergent too, thus it is not absolutely convergent.

So conclusion would be that it's conditionally convergent?
Would the absolute value of the fraction be $\frac{1}{n-1}$ or $\frac{1}{n+1}$ ? I wasn't quite sure about that.

Very good.