What is the sum from n=0 to infinity of (en)/(3n+1) ?

The sum from n=0 to infinity of (en)/(3n+1) is diverges

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sum from n=0 to infinity of (en)/(3n+1)

n = 0 \sum \infty \frac{e n}{3 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{e n}{3 n + 1}

Apply the constant multiplication rule:

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

= e \cdot n = 0 \sum \infty \frac{n}{3 n + 1}

Apply Series Divergence Test:

diverges

= e diverges

= diverges

\frac{6}{x}

t e^{- t^{5}}

\frac{\partial}{\partial x} ((x y)^{2})

\frac{d}{d x} (\frac{1}{x ^{1}})

\frac{d}{d x} (x e x)

What is the sum from n=0 to infinity of (en)/(3n+1) ?
The sum from n=0 to infinity of (en)/(3n+1) is diverges