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

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

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sum from n=1 to infinity of n/3

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= \frac{1}{3} diverges

= diverges

\frac{d}{d x} (\frac{3 x - 4}{6 x ^{2} - 1})

f (x) = ln (5 x - 7)

\int \frac{x ^{2} + 9}{x ^{2} + 4} d x

\int \frac{1}{7 y} d y

\frac{d y}{d x} = 1 - \frac{y}{x}

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