What is the sum from n=1 to infinity of (7n)/(5n+1) ?

The sum from n=1 to infinity of (7n)/(5n+1) is diverges

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sum from n=1 to infinity of (7n)/(5n+1)

n = 1 \sum \infty \frac{7 n}{5 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{7 n}{5 n + 1}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{n}{5 n + 1}

Apply Series Divergence Test:

diverges

= 7 diverges

= diverges

\int 20 x sec^{2} (x) d x

\frac{\partial}{\partial y} (\frac{x ^{2}}{1 + y ^{2}})

\frac{d}{d x} (- \frac{a}{x} + \frac{b}{x ^{2}})

- sin (2 x)

n = 0 \sum \infty 6

What is the sum from n=1 to infinity of (7n)/(5n+1) ?
The sum from n=1 to infinity of (7n)/(5n+1) is diverges