Determining whether a sequence of partial sums is convergent or divergent

**MathIsOhSoHard** · November 2nd 2012, 10:01 AM

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?

**Plato** · November 2nd 2012, 10:40 AM

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.

**MathIsOhSoHard** · November 2nd 2012, 11:56 AM

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)

**Plato** · November 2nd 2012, 12:31 PM

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 $\sum\limits_{n = 0}^\infty {ar^n }$

**MathIsOhSoHard** · November 2nd 2012, 02:07 PM

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

**Plato** · November 2nd 2012, 02:21 PM

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.

**MathIsOhSoHard** · November 2nd 2012, 03:02 PM

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 $(-1)^N$ ?

**Plato** · November 2nd 2012, 03:12 PM

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

**MathIsOhSoHard** · November 2nd 2012, 03:37 PM

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.

**Plato** · November 2nd 2012, 03:44 PM

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.

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