What is the sum from n=1 to infinity of (3e^n)/(n^5) ?

The sum from n=1 to infinity of (3e^n)/(n^5) is diverges

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sum from n=1 to infinity of (3e^n)/(n^5)

n = 1 \sum \infty \frac{3 e ^{n}}{n ^{5}}

diverges

Solution steps

n = 1 \sum \infty \frac{3 e ^{n}}{n ^{5}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{e ^{n}}{n ^{5}}

Apply Series Ratio Test:

diverges

= 3 diverges

= diverges

\frac{\partial}{\partial x} ((1 - 3 x y) e^{- x y})

\int (5^{x} - sec^{2} (x)) d x

\frac{\partial}{\partial y} (sin (3 x + yz))

\frac{\partial}{\partial x} (tan (3 x + 4 y))

\frac{x ^{2}}{x + 10}

What is the sum from n=1 to infinity of (3e^n)/(n^5) ?
The sum from n=1 to infinity of (3e^n)/(n^5) is diverges