What is the sum from n=1 to infinity of (2n)/(7n-2) ?

The sum from n=1 to infinity of (2n)/(7n-2) is diverges

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sum from n=1 to infinity of (2n)/(7n-2)

n = 1 \sum \infty \frac{2 n}{7 n - 2}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{7 n - 2}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{7 n - 2}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

\int \frac{1}{1 + cos ( x )} d x

x \to 0 lim (\frac{1}{x} + \frac{1}{x ^{2}})

n = 1 \sum \infty (2 n)!

n \to 0 lim (ln (n))

\int \frac{x}{x ^{2} + 8} d x

What is the sum from n=1 to infinity of (2n)/(7n-2) ?
The sum from n=1 to infinity of (2n)/(7n-2) is diverges