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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

x^{\frac{1}{2}}, 4

\frac{d}{d x} (2 x + \frac{π}{2})

x \to 0 + lim ((1 - x) sin (\frac{1}{x}))

\int 36 sec^{4} (x) d x

\int_{- 9}^{9} 1 - \frac{x ^{2}}{81} d x

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