# Determining whether a sequence of partial sums is convergent or divergent

• Nov 2nd 2012, 10:01 AM
MathIsOhSoHard
Determining whether a sequence of partial sums is convergent or divergent
Given the following sequence of partial sums:
$\displaystyle S_N=\frac{1}{N}cos(N\pi)$

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.

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

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

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

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

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

$\displaystyle a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$
Would this be correct?
• Nov 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:
$\displaystyle S_N=\frac{1}{N}cos(N\pi)$

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 $\displaystyle S_N=\frac{1}{N}cos(N\pi)$.
You know $\displaystyle (S_N)\to 0$ so the series converges to zero.

• Nov 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 $\displaystyle \sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $\displaystyle S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $\displaystyle a_n$ 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)
• Nov 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 $\displaystyle \sum^{/infty}_{n=1}a_n$ the sequence of partial sums is $\displaystyle S_N=\frac{1}{N}cos(N\pi)$. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
Hint: Find the series n'th term $\displaystyle a_n$ 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 $\displaystyle a_n$.
$\displaystyle a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

$$\sum\limits_{n = 0}^\infty {ar^n }$$ gives $\displaystyle \sum\limits_{n = 0}^\infty {ar^n }$
• Nov 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 $\displaystyle a_n$.
$\displaystyle a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

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

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

YES

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

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

How do you rewrite $\displaystyle (-1)^{N-1}$ into $\displaystyle (-1)^N$? :)
• Nov 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 $\displaystyle (-1)^{n-1}$ into $\displaystyle (-1)^n$? :)

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

You see $\displaystyle (-1)^{-1}=\frac{1}{-1}=-1$
• Nov 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:
$\displaystyle \left | a_n \right | \le \left | b_n \right |$
$\displaystyle \sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\displaystyle \sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $\displaystyle b_n$ 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 $\displaystyle \frac{1}{n-1}$ or $\displaystyle \frac{1}{n+1}$? I wasn't quite sure about that.
• Nov 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:
$\displaystyle \left | a_n \right | \le \left | b_n \right |$
$\displaystyle \sum^\infty_{n=1}\frac{1}{n}\le \sum^\infty_{n=1}(1)^n \left ( \frac{1}{n}+\frac{1}{n-1} \right )$

Since $\displaystyle \sum^\infty_{n=1}\frac{1}{n}$ is divergent, then the absolute of $\displaystyle b_n$ 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 $\displaystyle \frac{1}{n-1}$ or $\displaystyle \frac{1}{n+1}$? I wasn't quite sure about that.

Very good.