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

[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?

Thanks in advance

 

[checkitagain]

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

Thanks in advance

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.

 

[joao_pimentel]

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...

 

[checkitagain]

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...

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.

 

[joao_pimentel]

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"

 

[JJacquelin]

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"

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.

 

[joao_pimentel]

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

 

[JJacquelin]

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

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

 

[joao_pimentel]

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

 

[JJacquelin]

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.

 

[joao_pimentel]

God bless JJacquelin

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

Thank you very much for your work and for your kindness!

I read it carefully and I must confess I enjoyed this "Rice with geese à lá JJacquelin"

Thanks again, really