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

The sum from n=2 to infinity of 1/(2ie^nn^2) is converges

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sum from n=2 to infinity of 1/(2ie^nn^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

\frac{1}{2 i} : - \frac{i}{2}

= (- \frac{i}{2}) \cdot n = 2 \sum \infty \frac{1}{e ^{n} n ^{2}}

Apply Series Ratio Test:

converges

= (- \frac{i}{2}) converges

= converges

\int \frac{1}{e ^{a \cdot x}} d x

\int (csc (x) + tan (x))^{2} d x

x y^{^{'}} = y + 2 x^{2} y

\frac{d}{d x} ((1 + 5 x^{2}) (x - x^{2}))

What is the sum from n=2 to infinity of 1/(2ie^nn^2) ?
The sum from n=2 to infinity of 1/(2ie^nn^2) is converges