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

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

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

n = 1 \sum \infty \frac{5 n}{3 n - 1}

diverges

Solution steps

n = 1 \sum \infty \frac{5 n}{3 n - 1}

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

\frac{d}{d x} (- \frac{2 x}{x - 1})

f (x) = 7 (2 - 3 x)^{4}

\int \frac{( x ^{am} - x ^{n} )}{x} d x

y^{^{'}} + 3 y = 2 x e^{- 3 x}

\int 1 1^{x} d x

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