# Thread: Alternating Series - Converges or Diverges

1. ## Alternating Series - Converges or Diverges

Question:

Test the series for convergence or divergence.
$\displaystyle sn = \sum (-1)^n \frac{2n-1}{5n+1}$

My attempt:

This is an alternating series, therefore, I will use the alternating series test.

We know that sn+1 < sn, however, the limit of sn is $\displaystyle \frac{2}{5}$ and not 0. Therefore, this series diverges.

Am I correct?

Edit: I tried the ration test: $\displaystyle \frac{sn+1}{sn}$ and it showed that the series is convergent because $\displaystyle \lim \frac{sn+1}{sn} = 0$

2. Originally Posted by calc101
Question:

Test the series for convergence or divergence.
$\displaystyle s_n = \sum (-1)^n \frac{2n-1}{5n+1}$

My attempt:

This is an alternating series, therefore, I will use the alternating series test.

We know that sn+1 < sn, however, the limit of sn is $\displaystyle \frac{2}{5}$ and not 0. Therefore, this series diverges.

Am I correct?
That's correct!

3. Originally Posted by calc101
Edit: I tried the ration test: $\displaystyle \frac{sn+1}{sn}$ and it showed that the series is convergent because $\displaystyle \lim \frac{sn+1}{sn} = 0$
When you use the ratio test for absolute convergence, you have

$\displaystyle \lim\left|\left(-1\right)^n\frac{2n+1}{5n+6}\cdot\frac{5n+1}{2n-1}\right|=\lim\frac{\left(2n+1\right)\left(5n+1\ri ght)}{\left(5n+6\right)\left(2n-1\right)}=1$ which shows that the test is inconclusive.

4. We know that sn+1 < sn
When $\displaystyle n=1$, $\displaystyle s_1=-\frac{1}{6}$ but $\displaystyle s_{n+1}=s_2=-\frac{1}{6}+\frac{3}{11}$. Clearly, in this case, $\displaystyle s_{n+1}> s_n$.

Did you mean "$\displaystyle a_n$ is a decreasing sequence where $\displaystyle a_n= \left|(-1)^n \frac{2n-1}{5n+1} \right|=\frac{2n-1}{5n+1}$"?

5. Originally Posted by Showcase_22
When $\displaystyle n=1$, $\displaystyle s_1=-\frac{1}{6}$ but $\displaystyle s_{n+1}=s_2=-\frac{1}{6}+\frac{3}{11}$. Clearly, in this case, $\displaystyle s_{n+1}> s_n$.

Did you mean "$\displaystyle a_n$ is a decreasing sequence where $\displaystyle a_n= \left|(-1)^n \frac{2n-1}{5n+1} \right|=\frac{2n-1}{5n+1}$"?
I am sorry, I meant the sequence, an. Thank you for correcting me.

6. since the limit is 2/5 and not equal to zero, then the series is divergent.

7. Originally Posted by dr.tea
since the limit is 2/5 and not equal to zero, then the series is divergent.
Well of course $\displaystyle \lim _{n \to \infty } \left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}} \ne \frac{2}{5}$.

Here is the first test we all should apply.
The series $\displaystyle \sum {a_n }$ converges only if $\displaystyle \left( {a_n } \right) \to 0$.

So answer this question does $\displaystyle \left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}} \to 0 ?$

If not then $\displaystyle \sum {\left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}}}$ diverges.

8. The test of alternating series is
$\displaystyle \sum(-1)^{n}a_{n}$ converges if
1) $\displaystyle a_{n}$ is decreasing
2) \the limit of $\displaystyle a_{n}$ =0

since the second condition is not satisfied , then the sequence diverges
( lim $\displaystyle a_{n}$=2/5 )