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Evaluating the series $$f=\sum \frac{27^n}{(3n+1)!}$$ I could simplify this to $$\frac {1}{3}\sum \frac{3^{3n+1}}{(3n+1)!}$$ and use the chart $$\sum \frac{x^{an+b}}{(an+b)!}$$ to evaluate $f$. But The book hints at: (complex analysis)

As $1+\omega+\omega^2=0$ where $\omega$ is a complex root of unity,

$$e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$$

$$e^{\omega x}=\cdots$$

$$e^{\omega^2 x}=\cdots$$

$$e^x+\omega^2\cdot e^{\omega x}+\omega\cdot e^{\omega^2 x}=?$$ how are these related? It seems that I could obtain the series from this hint?

rebc
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1 Answers1

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If you look at the $(3n+1)!$ in the denominator, you will see that the series picks out every third summand of a related simple exponential expression.

The cube roots of unity come in, because if you follow the hint correctly, you will see that they pick out every third term for you. You fine tune the sum so that you pick out the $3n+1$ terms and not the $3n$ ones or the $3n+2$ ones.

You haven't mentioned in fact, that $\omega$ is a cube root of unity, but you will need to use that in your workings.

$$e^x=1+x+\frac {x^2}2+\frac {x^3}6+\dots$$

$$\omega^2e^{\omega x}=\omega^2+\omega^3x+\frac {\omega^4 x^2}2+\frac {\omega^5x^3}6+\dots$$

$$\omega e^{\omega^2 x}=\omega+\omega^3x+\frac {\omega^5 x^2}2+\frac {\omega^7x^3}6+\dots$$

Now use $\omega^3=1, 1+\omega+\omega^2=0$ and add these three expressions and see what happens, treating this as a power series in $x$. What will the next term be?

Mark Bennet
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