How do we evaluate the series $$\sum \frac{27^n}{(3n+1)!}$$ I could simplify this to $$\sum \frac{3^{3n}}{(3n+1)!}$$ but typically the tables provide you with the general series form of $$\sum \frac{x^{an+b}}{(an+b)!}$$ How do I go from herer? The b's are not the same!
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- 1$\begingroup$ See my answer to this question. $\endgroup$Lucian– Lucian2015-02-20 17:56:17 +00:00Commented Feb 20, 2015 at 17:56
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3 Answers
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1 You can write the series as $$\frac{1}{3}\sum \frac{3^{3n+1}}{(3n+1)!}$$right?
- $\begingroup$ hahah lol i guess youre right... $\endgroup$rebc– rebc2015-02-20 16:37:20 +00:00Commented Feb 20, 2015 at 16:37
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1 HINT:
As $1+\omega+\omega^2=0$ where $\omega$ is a complex root of unity,
$$e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$$
$$e^{\omega x}=\cdots$$
$$e^{\omega^2 x}=\cdots$$
$$e^x+\omega^2\cdot e^{\omega x}+\omega\cdot e^{\omega^2 x}=?$$
answered Feb 20, 2015 at 16:35
lab bhattacharjee
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- $\begingroup$ could you lead me more where were going with this? $\endgroup$rebc– rebc2015-02-20 21:16:18 +00:00Commented Feb 20, 2015 at 21:16
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3 Apply ratio test: $$\limsup_{n\to\infty}\left|\frac{27}{(3n+4)(3n+3)(3n+2)}\right|<1$$ So series converges by ratio test.
- $\begingroup$ This got downvoted in error. Could you do some editing so I can actually cancel my downvote and upvote it. Apologies $\endgroup$Shailesh– Shailesh2015-11-29 07:00:47 +00:00Commented Nov 29, 2015 at 7:00
- $\begingroup$ @Shailesh alright $\endgroup$user292378– user2923782015-11-29 08:02:03 +00:00Commented Nov 29, 2015 at 8:02
- $\begingroup$ Sorry it took time, but I canceled my downvote and upvoted it too. $\endgroup$Shailesh– Shailesh2015-12-19 12:53:56 +00:00Commented Dec 19, 2015 at 12:53