# Math Help - Alternating Series - Converges or Diverges

1. ## Alternating Series - Converges or Diverges

Question:

Test the series for convergence or divergence.
$
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 $\frac{2}{5}$ and not 0. Therefore, this series diverges.

Am I correct?

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

2. Originally Posted by calc101
Question:

Test the series for convergence or divergence.
$
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 $\frac{2}{5}$ and not 0. Therefore, this series diverges.

Am I correct?
That's correct!

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

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

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

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

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

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

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

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

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

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

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