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Check the series whether it is convergent /divergent?

$$\sum\limits_{n=0}^{\infty}\frac{(3i)^{2n+1}}{(2n+1)!}$$

I was thinking about the Taylor series but could not get its,,,,how to expand

I think the series is divergent by D'Alembert ratio test.

Am I right? Can you verify it and tell the solution, please? I would be grateful.

Thanks in advance.

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    $\begingroup$ I think you applied the ratio test in the wrong way, because the series is not only convergent but absolutely convergent. $\endgroup$ – Shashi Jan 4 '18 at 12:19
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    $\begingroup$ The expression is $\sinh(3i)$ $\endgroup$ – Thorgott Jan 4 '18 at 12:21
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Not only does your series converge by the ratio test (see the previous two answers), it can be summed.

As $$\sinh (z) = \sum_{n = 1}^\infty \frac{z^{2n + 1}}{(2n + 1)!}, \quad z \in \mathbb{C},$$ setting $z = 3i$ we have $$\sum_{n = 0}^\infty \frac{(3i)^{2n + 1}}{(2n + 1)!} = \sinh (3i) = i \sin (3).$$

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  • $\begingroup$ thanks a lot i like ur amswer @omegadot $\endgroup$ – user469754 Jan 6 '18 at 7:57
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Since$$\sum_{n=0}^\infty\left|\frac{(3i)^{2n+1}}{(2n+1)!}\right|=\sum_{n=0}^\infty\frac{3^{2n+1}}{(2n+1)!},$$which converges (by the ratio test), your series is absolutely convergent.

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The ratio test gives you

$$\frac{a_{n+1}}{a_n} = \frac{(3i)^{2n+3}(2n+1)!}{(2n+3)!(3i)^{2n+1}} = \frac{(3i)^2}{(2n+2)(2n+3)}$$

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  • $\begingroup$ thanks...@ 5xum $\endgroup$ – user469754 Jan 4 '18 at 12:32

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