# Dirichlet's Test on following sum

• May 17th 2010, 11:43 PM
jordanhe
Dirichlet's Test on following sum
Show that the series 1/(2log2)+1/(3log3)-1/(4log4)-1/(5log5)-1/(6log6)-1/(7log7)+.... is convergent, the rule of signs being that successive terms with the same sign come in groups of 2, 4, 8, 16... Begin by considering the series 1/2+1/3-1/4-1/5-1/6-1/7+1/8...

What i know is that the sum of 1/logn goes to 0 and decreasing, so that part is fine, but I can't find a bound on the 1/2+1/3-1/4-1/5-1/6-1/7+1/8... to satisfy the Dirichlet Test condition.
• May 18th 2010, 02:26 PM
chiph588@
The sum you're talking about is $\displaystyle \sum_{n=2}^\infty \frac{(-1)^{\lfloor \log_2(n)\rfloor-1}}{n}$, who's partial sums oscillate wildly between $\displaystyle 0$ and $\displaystyle 1$. It's hard to tell for inspection whether this sum converges or not but it does appear to be bounded.

What is this problem for? Is there any information you're leaving out?
• May 18th 2010, 02:40 PM
chiph588@
I think the sum I just mentioned diverges. See here.

The best way to show your original sum converges is by showing it converges absolutely. Can you take it from here?

Edit: The series is not absolutely convergent. It fails the integral test. Sorry for the mistake!
• May 18th 2010, 11:31 PM
chisigma
Quote:

Originally Posted by jordanhe
Show that the series 1/(2log2)+1/(3log3)-1/(4log4)-1/(5log5)-1/(6log6)-1/(7log7)+.... is convergent, the rule of signs being that successive terms with the same sign come in groups of 2, 4, 8, 16... Begin by considering the series 1/2+1/3-1/4-1/5-1/6-1/7+1/8...

What i know is that the sum of 1/logn goes to 0 and decreasing, so that part is fine, but I can't find a bound on the 1/2+1/3-1/4-1/5-1/6-1/7+1/8... to satisfy the Dirichlet Test condition.

Lets write...

$\displaystyle \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \dots = \sum_{n=1}^{\infty} (-1)^{n+1} a_{n}$ (1)

... where...

$\displaystyle a_{n} = \frac{1}{2^{n}} + \frac{1}{2^{n}+1} + \dots + \frac{1}{2^{n+1}-1}$ (2)

But $\displaystyle a_{n}<1$ and any partial sum $\displaystyle \sum_{n=1}^{N} (-1)^{n+1}$ is bounded, so that also any partial sum $\displaystyle \sum_{n=1}^{N} (-1)^{n+1} a_{n}$ is bounded...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 19th 2010, 04:51 AM
chiph588@
Quote:

Originally Posted by chisigma
Lets write...

$\displaystyle \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \dots = \sum_{n=1}^{\infty} (-1)^{n+1} a_{n}$ (1)

... where...

$\displaystyle a_{n} = \frac{1}{2^{n}} + \frac{1}{2^{n}+1} + \dots + \frac{1}{2^{n+1}-1}$ (2)

But $\displaystyle a_{n}<1$ and any partial sum $\displaystyle \sum_{n=1}^{N} (-1)^{n+1}$ is bounded, so that also any partial sum $\displaystyle \sum_{n=1}^{N} (-1)^{n+1} a_{n}$ is bounded...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Here's a similar approach:

If we look at the original sum and do what chisigma did we get the following:

$\displaystyle b_{n} = \frac{1}{2^{n}\log(2^n)} + \frac{1}{(2^{n}+1)\log(2^n+1)} + \dots + \frac{1}{(2^{n+1}-1)\log(2^{n+1}-1)} < \frac{a_n}{\log(2^n)}$ where $\displaystyle a_n$ is defined in the post above. Since $\displaystyle a_n<1$ we get that $\displaystyle b_n\to0$ so by the alternating series test our original sum converges since it equals $\displaystyle \sum (-1)^{n+1}b_n$.
• May 19th 2010, 07:51 AM
jordanhe
I think I know how to do it now. The thing is that we need to do the Dirichlet's Test twice. First, u use that on 1/2+1/3-1/4....., since the hypo-geometric series goes to zero and also decreasing, and the other one (1,1,-1,-1,-1,-1...) always bounded by zero, so this series is convergent. And then, since this is convergent, and 1/logn is decreasing and going to zero, so the original series will also be convergent.
• May 19th 2010, 08:24 AM
chiph588@
Quote:

Originally Posted by jordanhe
I think I know how to do it now. The thing is that we need to do the Dirichlet's Test twice. First, u use that on 1/2+1/3-1/4....., since the hypo-geometric series goes to zero and also decreasing, and the other one (1,1,-1,-1,-1,-1...) always bounded by zero, so this series is convergent. And then, since this is convergent, and 1/logn is decreasing and going to zero, so the original series will also be convergent.

That's the right idea, but keep in mind that first sum doesn't converge but is bounded and that's all we need.