Question

Show That
This power series
∑ n from 2 to infinity of
[ (-1)^{n} . z ^{2n+3} ] = [ (z^{7} ) / (1+z{^2} ) ] ; |z| <1

Is power series ∑ n from 2 to infinity of
[ (-1)^{n} . i ^ {2ni} . i ^(3i) ] Convergent ? If yes ,Compute It's Sum

Answer 1

1

Question #50128, Math, Complex Analysis

The formula $\sum_{n\geq 0}w^{n} = \frac{1}{1 - w}$, for $|z| < 1$ will be used for $w = -z^2$:

$$\sum_{n \geq 2} (-1)^n z^{2n+3} = z^7 (-1)^2 \sum_{n \geq 0} (-1)^n z^{2n} = z^7 \frac{1}{1 + z^2}$$

The power series $\sum_{n\geq 2}(-1)^n i^{2ni}i^{3i}$ has the terms

$$(-1)^n i^{2ni}i^{3i} = (-1)^n (e^{i\pi/2})^{2ni} (e^{i\pi/2})^{3i} = (-e^{-\pi})^n e^{-1.5\pi}$$

So, for $w = -e^{-\pi}$ and first formula

$$\sum_{n \geq 2} (-1)^n i^{2ni}i^{3i} = e^{-1.5\pi} e^{-2\pi} \sum_{n \geq 0} (-e^{-\pi})^n = e^{-3.5\pi} \sum_{n \geq 0} w^n = \frac{e^{-3.5\pi}}{1 - w} = \frac{e^{-3.5\pi}}{1 + e^{-\pi}} = \frac{e^{-2.5\pi}}{e^{\pi} + 1}$$

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