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I have this here power series which is complex valued

[tex]\frac{2n+1}{2^n} i^n = \frac{4}{25} + \frac{22}{25}i[/tex]

My task is to prove that this power series has the above mentioned sum.

To do this I seperate the sum into two sums

[tex]S_0 + S_1 = (i/2)^n + 2* (i/2)*n*(i/2)^{(n-1)}[/tex]

the first is easy since can use

[tex]S_0 = \frac{1}{1-(i/2)} = 4/5+2/5i[/tex]

The second is quite hard to compute. I'm to told that I need to differentiate, by using the sum

[tex]\sum _{n=0} ^{\infty} x^n = \frac{i}{2}[/tex]

then by differentiating the sum I get

[tex]f'(x) = \sum_{n=0} ^{\infty} n x^{n-1}[/tex]

In my textbook it says that x^n converge to 1/(1-x)

Then since |i/2| < 1, I differentiate both sides of the equation.

[tex]\frac{1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty} n (i/2)^{n-1}[/tex]

I multiply by i/2 on both sides of the equation to get.

[tex]\frac{i/2 * 1}{(1-(i/2))^2} =\sum_{n=0} ^{\infty}(i/2) n (i/2)^{n-1} [/tex]

then since |i/2| < 1, then I multiply by a factor of two on both sides of the eqation

[tex]2\frac{i/2 * 1}{(1-(i/2))^2} = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}[/tex]

[tex]-16/25 + 12/25i = \sum_{n=0} ^{\infty}2 (i/2) n (i/2)^{n-1}[/tex]

Finally [tex]\frac{4}{5}+ \frac{2}{5i}+ (\frac{-16}{25} + \frac{12}{25i}) = \frac{4}{25} + \frac{22}{25}i[/tex]

I have the feeling that I'm doing something wrong?

Sincerely

Hummingbird25

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# Homework Help: HELP: A complex valued power series

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