# How do you find a formula for the sum n terms Sigma (1+(2i)/n)^2(2/n) and then find the limit as n->oo?

Oct 26, 2016

The sum diverges.

#### Explanation:

${\left(1 + \frac{2 i}{n}\right)}^{2} \left(\frac{2}{n}\right) = \left(1 + 2 i \left(\frac{2}{n}\right) - {\left(\frac{2}{n}\right)}^{2}\right) \left(\frac{2}{n}\right) =$
$\frac{2}{n} + 2 i {\left(\frac{2}{n}\right)}^{2} - {\left(\frac{2}{n}\right)}^{3}$

then

${\sum}_{n = 1}^{\infty} {\left(1 + \frac{2 i}{n}\right)}^{2} \left(\frac{2}{n}\right) = 2 {\sum}_{n = 1}^{\infty} \frac{1}{n} + 8 i {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 - 8 {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 3$

Here

${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges
${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$
${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 3 = \zeta \left(3\right) \approx 1.20206$

so the sum

${\sum}_{n = 1}^{\infty} {\left(1 + \frac{2 i}{n}\right)}^{2} \left(\frac{2}{n}\right)$ diverges