# Complex Infinite Series

Having trouble with this infinite series and deciding whether it converges or diverges.

The series:

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

My thoughts are that you take the modulus of the fraction and get $$\frac{1}{2}$$ to the exponent $$n$$ makes it go to $$0$$ and then multiplied by $$n$$ make it

$$\infty*0$$ which is always divergent right, making the series diverge? Can someone also clarify that this is the case?

• No this is not true, the summation of $n \cdot r^n$ converges where $|r| \lt 1$ – Peter Foreman Feb 2 at 16:45

First let’s look if the series converges absolutely.

For this, we need to see if $$\sum b_n = \sum \frac{n}{2^n}$$ converges. And this is immediate using the ratio test

as $$\lim\limits_{n\to \infty}\frac{b_{n+1}}{b_n} =1/2<1$$.

Conclusion: the given series converges absolutely hence converges

• Limit 0 in the ratio test, sure about that? – Did Feb 2 at 16:51
• @Did Thanks for asking the question! – mathcounterexamples.net Feb 2 at 16:53

Hint. Your first thought is correct: look at the modulus.

Your reasoning about $$\infty * 0$$ is wrong.

Try the ratio test.

If you know about the geometric series $$1 + x + x^2 + \cdots$$ you can differentiate, multiply by $$x$$ and actually find out what your series converges to.

Consider $$\sum_{n=0}^\infty nz^n$$. The radius of convergence is $$r=\limsup_{n\to\infty}\frac1{n^{\frac1n}}=1$$. Since $$\mid\frac1{2i}\mid=\frac12$$, the series converges.