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| sum 1/(2^k+3^k) https://forumdematematica.org/viewtopic.php?f=11&t=112 |
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| Autor: | kinu [ 01 jan 2012, 10:50 ] |
| Título da Pergunta: | sum 1/(2^k+3^k) |
\(\displaystyle{\sum_{k=0}^{n}\frac{1}{2^k+3^k}=}\) |
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| Autor: | João P. Ferreira [ 02 jan 2012, 01:15 ] |
| Título da Pergunta: | Re: sum |
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' |
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| Autor: | João P. Ferreira [ 04 jan 2012, 12:33 ] | ||
| Título da Pergunta: | Re: sum | ||
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
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| Autor: | kinu [ 18 jan 2012, 13:55 ] |
| Título da Pergunta: | Re: sum |
Thanks Admin for Nice solution |
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