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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?

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

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

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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.

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