# Series convergence question

• Jan 27th 2011, 01:40 AM
Glitch
Series convergence question
The question:

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

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

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

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

Thus the series converges.

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

Using ratio test:

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

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

Where have I gone wrong? Thanks.
• Jan 27th 2011, 01:48 AM
Ted
First, in ratio test you'll calculate the limit of the fraction $\displaystyle \dfrac{a_{k+1}}{a_k}$ 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.
• Jan 27th 2011, 01:51 AM
FernandoRevilla
Quote:

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

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

Fernando Revilla
• Jan 27th 2011, 02:02 AM
FernandoRevilla
Quote:

Originally Posted by Ted
First, in ratio test you'll calculate the limit of the fraction $\displaystyle \dfrac{a_{k+1}}{a_k}$ 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 $\displaystyle a_k$ satisfies $\displaystyle a_{k+1}/a_k\leq r$ for $\displaystyle k$ sufficiently large and $\displaystyle 0\leq r <1$ fixed then , the series is convergent. In our case does not exist such $\displaystyle r$ .

Fernando Revilla
• Jan 27th 2011, 02:04 AM
Ted
Well, am talking about the standard ratio test :p .
• Jan 27th 2011, 03:04 AM
chisigma
Dirichlet's Test -- from Wolfram MathWorld

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 27th 2011, 03:25 AM
Plato
Quote:

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

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

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

The above is correct. The given series converges.

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

It easy to prove that if $\displaystyle \alpha>0$ then $\displaystyle \sum\limits_{k = 3}^\infty \dfrac{1}{[ln(k)]^{\alpha}}$ diverges.
• Jan 27th 2011, 04:30 AM
Glitch
This is what I've tried:

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

$\displaystyle b_k \le a_k$

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

= $\displaystyle ln(ln(n + 1)) - ln(ln(3)) \to \infty$ as $\displaystyle n \to \infty$, therefore $\displaystyle a_k$ must diverge.
• Jan 27th 2011, 04:49 AM
FernandoRevilla
Better:

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

and

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

is divergent.

Fernando Revilla