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

2 x, 2 x - 4, 0, 4

\int - 2 t^{2} d t

\frac{\partial}{\partial x} (cos (4 x + 5 y) \cdot 5)

\frac{5}{2 s ^{2} + 3 s + 1}

\frac{d}{d x} (0.3)

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