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Problem

Investigate whether the sum $$\sum_{n=1}^{\infty}e^{\frac{i\pi n}{2}} \cdot \frac{1}{n^2}$$

is convergent.

My attempt

The comparison theorem states that for $0\leq a_n \leq b_n$ for $n \in \mathbb{N}$ the following holds:

  1. If $\sum_{n=1}^{\infty}b_n $ is convergent, then $\sum_{n=1}^{\infty}a_n $ is also convergent.
  2. If $\sum_{n=1}^{\infty}a_n $ is divergent, then $\sum_{n=1}^{\infty}b_n $ is also divergent

Let's call $b_n=e^{\frac{i\pi n}{2}}$ and $a_n=e^{\frac{i\pi n}{2}} \cdot \frac{1}{n^2}$. I believe we now meet the condition that $0 \leq a_n \leq b_n$.

And from here I don't know how to get further. Have I even used the correct method or am I just making it harder for myself. Because if $a_n$ really is convergent to prove it, I first have to show that $b_n$ is convergent, which might not be as easy as I thought.

Can anyone help me out?

Carl
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    $e^{\frac{i\pi n}{2}}$ isn't real for odd $n$ and can be negative when $n \equiv 2\pmod 4$. You should use "absolute convergence $\implies$ convergence" instead. – achille hui Sep 21 '20 at 11:24

1 Answers1

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We have that the series converges by absolute convergence test, indeed

$$\sum_{n=1}^{\infty}\left|e^{\frac{i\pi n}{2}} \cdot \frac{1}{n^2}\right|=\sum_{n=1}^{\infty}\frac{1}{n^2} =\frac{\pi^2}6$$

Refer also to the related

user
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  • Thanks for the quick answer. But, I'm not sure why $e^{\frac{i \pi n}{2}}$ vanishes when we take the absolute value of the entire expression? Can you clarify it for me? @user – Carl Sep 21 '20 at 11:45
  • More precisely we are taking the modulus and by that we can conclude that both real and imaginary part converge therefore the given series converges. Refer also to this. – user Sep 21 '20 at 11:55
  • Well, it seems like it's very clear to you, but I really don't understand how $\sum_{n=1}^{\infty}\left|e^{\frac{i\pi n}{2}} \cdot \frac{1}{n^2}\right|=\sum_{n=1}^{\infty}\frac{1}{n^2}$. Are you skipping some steps? @user – Carl Sep 21 '20 at 12:32
  • @Carl Recall that $|z_1 \cdot z_2|= |z_1 \cdot |z_2|$ therefore $$\left|e^{\frac{i\pi n}{2}} \cdot \frac{1}{n^2}\right|=\left|e^{\frac{i\pi n}{2}} \right|\cdot \left| \frac{1}{n^2}\right|=\frac{1}{n^2}$$ – user Sep 21 '20 at 12:36