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

1. ## 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?

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

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.

3. ## 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)

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

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 $a_n$.
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

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

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

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 $a_n$.
$a_N=S_{N}-S_{N-1}=(-1)^N\left[\frac{2N-1}{N(N-1)}\right]$.

$$\sum\limits_{n = 0}^\infty {ar^n }$$ gives $\sum\limits_{n = 0}^\infty {ar^n }$
Are you sure $a_n$ can't be written as $a_N=(-1)^N\cdot \left ( \frac{1}{N}-\frac{1}{N-1}\right )$

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

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

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

Originally Posted by Plato
YES

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

How do you rewrite $(-1)^{N-1}$ into $(-1)^N$?

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

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

9. ## 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.

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

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.