sum 1/(2^k+3^k)

[kinu]

\(\displaystyle{\sum_{k=0}^{n}\frac{1}{2^k+3^k}=}\)

[João P. Ferreira]

What shall I suppose to find? The general expression?

We can clearly see that it converges

\(\frac{1}{3^k+3^k}<\frac{1}{2^k+3^k}<\frac{1}{2^k+2^k}\)

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

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

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

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

So it converges

But I'm not really seeing how shall I calculate the general expression... but if you find out kindly let me know..

regards

PS: And next time dear kinu just try to use the word 'please'

[João P. Ferreira]

Hi

Thanks to a very kind user named JJacquelin the solution was found

The q-polygamma functions (do not confuse with the usual polygamma functions) were used

See the image

Best regards

Anexos

q-digamma.JPG (43.67 KiB) Visualizado 4688 vezes

[kinu]

Thanks Admin for Nice solution