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I have to calculate the series $$ S=\sum_{ n=1}^{\infty} {\frac{{\rm i}^{n}}{3^{n}}} $$

  • I know that it's result is $S = \left(-1 + 3{\rm i}\right)/10$.
  • I have tried to think in using alternatives methods to solve this problem.
  • I was thinking of multiplying by ${\rm i}^{2}$ to the sum, with the intention of having $S - {\rm i}^{2}S = 0$.
  • Multiplying by ${\rm i}^{2}$ would not destroy the geometric properties of the series, but it would give me $S = 0$ as result which it's wrong.

I don't see where the mistake is.

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    $\begingroup$ Why is $S-i^{2}S=0$? $\endgroup$ Commented Jun 24 at 11:41
  • $\begingroup$ $i^2S=\sum_{n=1}^{\infty} \frac{i^{n+2}}{3^n}=-S\ne S$ (if $S\ne 0$ of course). If think you meant $S-i^4S=0$, but I don't know if it will help in any way. $\endgroup$ Commented Jun 24 at 11:48
  • $\begingroup$ The idea behind the proposed approach is to multiply by a number $r \neq 1$ in a way that $S$ and $rS$ have all but finitely many terms the same. The "usual choice" is the ratio of consecutive terms, $r = i/3$. To find an alternative along the same lines, you presumably want a finite sequence of algebraic operations that "effectively cancel" all but finitely many terms (but not all of them). $\endgroup$ Commented Jun 24 at 11:54

3 Answers 3

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Using $$\frac{i^{2n}}{3^{2n}}=\frac{(-1)^n}{9^n}$$ you have

$$\sum _{n=1}^{\infty } \frac{i^n}{3^n}=\sum _{n=0}^{\infty } \frac{i^{2 n+1}}{3^{2 n+1}}+\sum _{n=1}^{\infty } \frac{i^{2 n}}{3^{2 n}}=\sum _{n=1}^{\infty } \frac{i^{2 n}}{3^{2 n}}-3 i \sum _{n=1}^{\infty } \frac{i^{2 n}}{3^{2 n}}=(1-3 i) \sum _{n=1}^{\infty } \left(-\frac{1}{9}\right)^n=-\frac{1}{10}+\frac{3 i}{10}$$

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As mention in a comment, multiplying by $i^2$ won't help because $i^2=-1$.

One way to prove this identity (if we assume we only know real coefficient geometric series) is to separate odd terms from even terms: $$\sum_{n=1}^{+\infty}\frac{i^n}{3^n}=\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{9^n}+\frac{i}{3}\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{9^n}=-1+\left(1+\frac{i}{3}\right)\sum_{n=0}^{+\infty}\left(\frac{-1}{9}\right)^n=-1+\left(1+\frac{i}{3}\right)\frac{9}{10}=\frac{-1+3i}{10}$$

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Alternative approach:

In real analysis, you have that for $~| ~x ~| < 1,~$ that

$$\sum_{i=0}^\infty x^i = \frac{1}{1 - x}.\tag1 $$

The formula in (1) above is based on two ideas:

  • $\displaystyle \sum_{i=0}^n x^i = \frac{1 - x^{n+1}}{1 - x}.$

  • $\displaystyle |x| < 1 \implies \lim_{n\to\infty} x^n = 0.$

The entire conception above translates directly into complex analysis, because:

  • The algebra of the first bullet point above works just as well for $~z \neq 1 ~: ~z \in \Bbb{C}.~$

  • For any $~z \in \Bbb{C} ~: | ~z ~| < 1,~$ you (also) have that $~\displaystyle \lim_{n\to\infty} z^n = 0.$

Therefore, you have that

$$\sum_{i=0}^\infty z^i = \frac{1}{1 - z} ~: ~z \in \Bbb{C}, ~| ~z ~| < 1.\tag2 $$

With the present problem, setting $~z = \dfrac{i}{3} \implies ~| ~z ~| < 1.$

Then, you have that

$$\sum_{i=1}^\infty z^i = z \times \sum_{i=0}^\infty z^i$$

$$= z \times \frac{1}{1 - z} = \frac{i}{3} \times \frac{1}{1 - \frac{i}{3}}$$

$$= \frac{i}{3} \times \frac{3}{3-i} \times \frac{3 + i}{3+i}$$

$$= \frac{i}{3} \times \frac{9 + 3i}{10}$$

$$= i \times \frac{3 + i}{10} = \frac{-1 + 3i}{10}.$$

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