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## It's time for a Series Contest!

Moderator: galactus

### It's time for a Series Contest!

Wed May 14, 2014 10:09 am

Posts: 850
Location: Jaipur, India

It's time for a Series Contest! What do you say guys?

### Re: It's time for a Series Contest!

Wed May 14, 2014 11:47 am
galactus
Global Moderator

Posts: 902
whatever you want, S. I'm easy to get along with

sounds fun.

Here's one to start:

$$\displaystyle \sum_{k=1}^{\infty}Ci(2\pi nk)=1/2\left(\psi(n)-log(n)+\frac{1}{2n}\right)$$

as you know what Ci is, the Cosine integral defined by $$-\int_{x}^{\infty}\frac{\cos(t)}{t}dt=Ci(x)$$

I want to take a look at that 'famous' integral you just posted. But, I have to leave for a while. I will look at it this evening.

Some clever use of contours may be a fun way to do this one. Maybe not the easiest...but fun. Maybe a Hankel contour?. Just a thought I have for now from looking at it.

### Re: It's time for a Series Contest!

Wed May 14, 2014 1:01 pm

Posts: 850
Location: Jaipur, India

galactus wrote:
whatever you want, S. I'm easy to get along with

sounds fun.

Here's one to start:

$$\displaystyle \sum_{k=1}^{\infty}Ci(2\pi nk)=1/2\left(\psi(n)-log(n)+\frac{1}{2n}\right)$$

as you know what Ci is, the Cosine integral defined by $$-\int_{x}^{\infty}\frac{\cos(t)}{t}dt=Ci(x)$$

I want to take a look at that 'famous' integral you just posted. But, I have to leave for a while. I will look at it this evening.

Some clever use of contours may be a fun way to do this one. Maybe not the easiest...but fun. Maybe a Hankel contour?. Just a thought I have for now from looking at it.

I am planning to start up tomorrow. It's good for increasing forum activity. Of course the first problem will be the one which you suggest.

### Re: It's time for a Series Contest!

Thu May 15, 2014 6:15 am