# Find the general expression from k=0 to 'n'

1. Jan 1, 2012

### joao_pimentel

Hi guys... this problem is really annoying me

How to find:
$$\displaystyle{\sum_{k=0}^{n}\frac{1}{2^k+3^k}}$$

I can clearly see that it converges

$$\frac{1}{3^k+3^k}<\frac{1}{2^k+3^k}<\frac{1}{2^k+2^k}$$

$$\sum_{k=0}^{\infty}\frac{1}{3^k+3^k}<\sum_{k=0}^{\infty}\frac{1}{2^k+3^k}<\sum_{k=0}^{\infty}\frac{1}{2^k+2^k}$$

$$\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{3^k}<\sum_{k=0}^{\infty}\frac{1}{2^k+3^k}<\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{2^k}$$

$$\frac{1}{2}\frac{1}{1-1/3}<\sum_{k=0}^{\infty}\frac{1}{2^k+3^k}<\frac{1}{2}\frac{1}{1-1/2}$$

$$\frac{3}{4}<\sum_{k=0}^{\infty}\frac{1}{2^k+3^k}<1$$

But how do I find the general expression from k=0 to 'n' or even the exact value as n goes to infinity?

2. Jan 1, 2012

### checkitagain

Re: sum

You must be assuming that it does. I bet there is no closed form.
It could be a "wild goose chase."

So, the origin of the problem and/or who presented it
need to be considered.

3. Jan 2, 2012

### joao_pimentel

Re: sum

What do you mean by "wild goose chase."?

I was wondering if is there any way of separating $\frac{1}{2^k+3^k}$ in two or more simple fractions...

4. Jan 2, 2012

### checkitagain

Re: sum

Here might be the Devil's advocate:

It would be a wild goose chase, in fact, there is not a solution.
Looking for something that turns out to not to be there
would be a wild goose chase. It is for the consideration of
users here not working on a problem if there is not a definite solution,
but that there was supposed to be.

Last edited: Jan 2, 2012
5. Jan 3, 2012

### joao_pimentel

Re: sum

OK but I must confess that I love wild goose hunting, so I was wondering if someone could just give me a tip on how shall I load the rifle. And if I hit the goose, I'll seve you with one of the best and popular portuguese dishes "Rice with geese"

6. Jan 3, 2012

### JJacquelin

Re: sum

The size of rifle should be not for goose hunting, but rather for elephant and better for dragon !
You have a small chance to find a closed form for this kind of infinite series if you are very smart in using special functions and especially if you have a perfect knowledge of the q-polygamma functions (do not confuse with the usual polygamma functions).
Nevertheless, I am quite certain that the closed form can be expressed in terms of q-polygamma functions.
But instead of spending a lot of time in doing it, I have the pleasure to let the work to be done by somone else more available and more smart than myself.

7. Jan 3, 2012

### joao_pimentel

Re: sum

I'm sorry my alleged arrogance, it was never my intention to state that I'm more smart than anyone, I was just trying to find a solution to a problem which I found thrilling.
Q-polygamma functions might be the answer.

Thank you very much for your kind attention

8. Jan 3, 2012

### JJacquelin

Re: sum

I never intended to call someone arrogant.
Simply, I was in the mood to joke.
Have a nice time with the q-polygamma ! :rofl:

9. Jan 3, 2012

### joao_pimentel

Re: sum

Thank you very much... For sure I'll have :)

10. Jan 4, 2012

### JJacquelin

Re: sum

Hello joao_pimentel !

finally, I went to the goose hunting.
With the invavuable help of WolframAlpha for the most boring part of job, I don't comme back with empty bag.
In the joint page, find the formulas for finite and infinite series with general term 1/(a^k+b^k).
Numerical verification in case a=3, b=2 . (More digits could be provided, thanks to Wolframalpha). Of course, the direct numerical computation of the sum is much easier than with these complicated formulas.

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11. Jan 4, 2012

### joao_pimentel

Re: sum

God bless JJacquelin

Outstanding results I must confess, for sure my rifle would never be enough powerful to shot down this dragon.