# Math Help - Series convergence question

1. ## Series convergence question

The question:

Determine if the following alternating series converges. Is it absolutely convergent?

$\sum\limits_{k = 3}^\infty \frac{(-1)^k}{ln(k)}$

My attempt:
Using the alternating series test, we need:
a) $a_k \ge 0$
b) $a_{k} \ge a_{k + 1}$
c) $\lim\limits_{k \to \infty} a_k = 0$

a) $\frac{1}{ln(k)}$ > 0 for all k > 1
b) $\frac{1}{ln(k)} \ge \frac{1}{ln(k+1)}$ for all k > 1
c) Limit is obviously 0 as $k \to \infty$

Thus the series converges.

Now we consider $\sum\limits_{k = 3}^\infty |\frac{1}{ln(k)}|$

Using ratio test:

$r = \frac{a_{k + 1}}{a_k} = |\frac{1}{ln(k + 1)}}| |\frac{ln(k)}{1}}|$
= $\frac{ln(k)}{ln(k+1)} < 1$

Therefore converges. So this series should be absolutely convergent. However the answer is actually 'conditionally convergent'.

Where have I gone wrong? Thanks.

2. First, in ratio test you'll calculate the limit of the fraction $\dfrac{a_{k+1}}{a_k}$ as k goes to infinity not just the fraction!
Second, the ratio test will be failed.
Third, I'll use the comparison test for the absolute series.

3. Originally Posted by Glitch
Where have I gone wrong? Thanks.

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{\log k}{\log (k+1)}}=1$

Fernando Revilla

4. Originally Posted by Ted
First, in ratio test you'll calculate the limit of the fraction $\dfrac{a_{k+1}}{a_k}$ as k goes to infinity not just the fraction!
There is a ratio test without calculating the limit:

If the series of general positive term $a_k$ satisfies $a_{k+1}/a_k\leq r$ for $k$ sufficiently large and $0\leq r <1$ fixed then , the series is convergent. In our case does not exist such $r$ .

Fernando Revilla

5. Well, am talking about the standard ratio test .

6. Dirichlet's Test -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

7. Originally Posted by Glitch
Determine if the following alternating series converges. Is it absolutely convergent?
$\sum\limits_{k = 3}^\infty \frac{(-1)^k}{ln(k)}$

Using the alternating series test, we need:
a) $a_k \ge 0$
b) $a_{k} \ge a_{k + 1}$
c) $\lim\limits_{k \to \infty} a_k = 0$

a) $\frac{1}{ln(k)}$ > 0 for all k > 1
b) $\frac{1}{ln(k)} \ge \frac{1}{ln(k+1)}$ for all k > 1
c) Limit is obviously 0 as $k \to \infty$
Thus the series converges.
The above is correct. The given series converges.

BUT $\sum\limits_{k = 3}^\infty \left|\dfrac{(-1)^k}{ln(k)}\right|=\sum\limits_{k = 3}^\infty \dfrac{1}{ln(k)}$.

It easy to prove that if $\alpha>0$ then $\sum\limits_{k = 3}^\infty \dfrac{1}{[ln(k)]^{\alpha}}$ diverges.

8. This is what I've tried:

Let $a_k = \abs{\frac{1}{ln(k)}}$
Let $b_k = \abs{\frac{1}{kln(k)}}$

$b_k \le a_k$

$\int\limits_{3}^{n} \frac{1}{kln(k)} \le \sum\limits_{k = 3}^\infty \frac{1}{ln(k)}$

= $ln(ln(n + 1)) - ln(ln(3)) \to \infty$ as $n \to \infty$, therefore $a_k$ must diverge.

9. Better:

$\displaystyle\lim_{k \to{+}\infty}{\dfrac{1/\log k}{1/k}}=+\infty$

and

$\displaystyle\sum_{k=3}^{+\infty}\dfrac{1}{k}$

is divergent.

Fernando Revilla