burkinaval1b

2022-01-04

Having trouble with this infinite series and deciding whether it converges or diverges.
The series:
$\sum _{n=1}^{\mathrm{\infty }}n{\left(\frac{1}{2i}\right)}^{n}$

Bubich13

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 $\underset{n\to \mathrm{\infty }}{lim}\frac{{b}_{n+1}}{{b}_{n}}=\frac{1}{2}<1$
Conclusion: the given series converges absolutely hence converges

Chanell Sanborn

Hint. Your first thought is correct: look at the modulus.
Your reasoning about $\mathrm{\infty }\cdot 0$ is wrong.
Try the ratio test.
If you know about the geometric series
$1+x+{x}^{2}+\dots$
you can differentiate, multiply by x and actually find out what your series converges to.

karton

Consider $\sum _{n=0}^{\mathrm{\infty }}n{z}^{n}$. The radius of convergence is $r=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{\frac{1}{n}}}=1$. Since $|\frac{1}{2i}|=\frac{1}{2}$, the series converges.