# Does $\sum_{n\geq 1}\left(\frac{1+2i}{\sqrt{5}}\right)^n$ converge?

I am having trouble understanding how to apply convergence tests to complex series. Which test(s) could I use here to determine whether this series converges or diverges? Thanks!

• This one has $r=|(1+2i)/\sqrt 5| =1$, and it has a bounded partial sum. However, this does not converge. Try to imitate the proof of convergence of geometric series, and use it to find a formula for partial sum. – Sungjin Kim May 24 '17 at 16:55

It's a geometric series $\sum_n r^n$ with $r = (1+2i)/\sqrt{5}$. It is guaranteed to converge if $|r|<1$ and diverge if $|r|>1$.

Check if $$|r| = \left|\frac{1+2i}{\sqrt{5}}\right| < 1.$$

I initially wanted to leave this as a hint, however I'll just go into greater detail. As discovered, you get that $|r|=1$ exactly. Hence the terms $$\left|\frac{1+2i}{\sqrt{5}}\right|^n = 1 \not\to 0$$ as $n\to\infty$, hence the series diverges by the $n$th term test.

• The absolute value of $1+2i$ is $\sqrt{5}$ so this makes $|r|=1$. What does this mean? – MathStudent1324 May 24 '17 at 16:52
• That it does not converge, as the general term does not even go to $0$. – Clement C. May 24 '17 at 16:53
• I also know that the series $\sum_{n\geq 1}(1)^n$ diverges, but in this series, does the fact that there is the $2i$ affect this? Specifically, I would think that $i$ when raised to different powers might cancel out some terms. – MathStudent1324 May 24 '17 at 16:54
• No, no such cancellations can help. For every $n$, $\lvert (e^{i t})^n\rvert = 1 \xrightarrow[n\to\infty]{} 0$. A series whose general term does not converge to $0$ trivially diverges. – Clement C. May 24 '17 at 16:55

$z=\frac{1+2i}{\sqrt{5}}$ is a point on the unit circle, $z=e^{i\arctan 2}$. It follows that $$S_N = \sum_{n=1}^{N} z^n = z\cdot\frac{z^N-1}{z-1}$$ has a bounded modulus, but it is not converging as $N\to +\infty$.

$$\frac{1+2i}{\sqrt 5} = \cos\varphi + i \sin \varphi \text{ where } \varphi = \arctan 2.$$ So $$\left( \frac{1+2i}{\sqrt 5} \right)^n = \cos(n\varphi) + i\sin(n\varphi).$$ The terms remain on the circle $|z|=1$ as $n$ changes. Therefore the terms do not approach $0.$ Therefore the series diverges.

(However it does have a Cesàro sum.)