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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{0.5}}

\frac{d}{d t} (4 e^{- 3 t})

\frac{d}{d A} (- cos (2) (A) - sin (2) (A))

\int \frac{1}{x + y} d x

f (x) = \frac{2 x ^{2} + 4 x + 2}{x}

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