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I have come up with a solution but I am not sure if it is correct. Here is what I have so far:

$$\sum_{n = 1}^{\infty} \frac{(2-i)}{(1-i)^n}$$ $$= \sum_{n = 1}^{\infty} \frac{2}{(1-i)^n} - \sum_{n = 1}^{\infty} \frac{i}{(1-i)^n}$$

Now, let's assume that $S_n$ is the partial sum of the first $n$ terms of the series: $$ \sum_{n = 1}^{\infty} \frac{2}{(1-i)^n}$$ Then:

$$S_1 = \frac{2}{(1-i)}$$ $$S_2 = \frac{2}{(1-i)} + \frac{2}{(1-i)^2} = \frac{2(1-i)+2}{(1-i)^2}$$ $$S_3 = \frac{2}{(1-i)} + \frac{2}{(1-i)^2} + \frac{2}{(1-i)^3} = \frac{2(1-i)^2 + 2(1-i) + 2}{(1-i)^3}$$ I can see that the pattern is: When $n \geq 2$: $$S_n = \frac{2(1-i)^{n-1} + 2(1-i)^{n-2} + ... + 2(1-i) + 2}{(1-i)^n} $$ When $n = 1 $: $$S_1 = \frac{2}{(1-i)}$$

For the series: $$\sum_{n = 1}^{\infty} \frac{i}{(1-i)^n}$$ the same concept applies, as such: When $n \geq 2$: $$S_n = \frac{i(1-i)^{n-1} + i(1-i)^{n-2} + ... + i(1-i) + i}{(1-i)^n} $$ When $n = 1 $: $$S_1 = \frac{i}{(1-i)}$$

So, we have:

$$= \sum_{n = 1}^{\infty} \frac{2}{(1-i)^n} - \sum_{n = 1}^{\infty} \frac{i}{(1-i)^n}$$ $$= \frac{2}{(1-i)} + \frac{2(1-i)^{n-1} + 2(1-i)^{n-2} + ... + 2(1-i) + 2}{(1-i)^n} - \frac{i}{(1-i)} - \frac{i(1-i)^{n-1} + i(1-i)^{n-2} + ... + i(1-i) + i}{(1-i)^n} $$ (For the limits, I will ignore the lesser order terms.) $$ = \frac{2}{(1-i)} + \lim_{n \rightarrow \infty}{\frac{2(1-i)^{n-1}}{(1-i)^n}}- \frac{i}{1-i} - \lim_{n \rightarrow \infty}{\frac{i(1-i)^{n-1}}{(1-i)^n}}$$ $$= \frac{2}{1-i} + \frac{2}{1-i} - \frac{i}{1-i} - \frac{i}{1-i}$$ (From complex division) $$ = \frac{6+2i}{2}$$ $$= 3 + i$$

Is this correct or have I made a mistake somewhere?

Thanks

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    http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+(2-i)%2F(1-i)%5En – 2017-02-15

1 Answers 1

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You have made things WAY too hard for yourself.

This is a geometric series, with initial term $a=\frac{2-i}{1-i}$ and common ratio $r=\frac{1}{1-i}$. So, as long as $\lvert r\rvert<1$, we know that the series will converge to $\frac{a}{1-r}$.

In this case, $$ \lvert r\rvert=\left\lvert\frac{1}{1-i}\right\rvert=\left\lvert\frac{1+i}{2}\right\rvert=\frac{\sqrt{2}}{2}<1. $$ So, we know that the series converges to $$ \begin{align*} \frac{a}{1-r}&=\frac{\left(\frac{2-i}{1-i}\right)}{1-\frac{1}{1-i}}\\ &=\frac{2-i}{1-i-1}\\ &=\frac{2-i}{-i}\\ &=1+2i. \end{align*} $$