What is the sum from n=1 to infinity of (3n^2)/(2i) ?

The sum from n=1 to infinity of (3n^2)/(2i) is diverges

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sum from n=1 to infinity of (3n^2)/(2i)

n = 1 \sum \infty \frac{3 n ^{2}}{2 i}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n ^{2}}{2 i}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{3}{2 i} \cdot n = 1 \sum \infty n^{2}

Apply Series Divergence Test:

diverges

= \frac{3}{2 i} diverges

= diverges

n = 1 \sum \infty \frac{n}{5 n}

n = 0 \sum \infty (\frac{2 - 1}{2})^{n}

n = 1 \sum \infty (- 1)^{n} 0

n = 4 \sum \infty \frac{4}{( n ^{2} - 1 )}

n = 1 \sum \infty \frac{n + 8}{n ^{2}}

What is the sum from n=1 to infinity of (3n^2)/(2i) ?
The sum from n=1 to infinity of (3n^2)/(2i) is diverges