What is the sum from n=1 to infinity of (2n)/(7n-6) ?

The sum from n=1 to infinity of (2n)/(7n-6) is diverges

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sum from n=1 to infinity of (2n)/(7n-6)

n = 1 \sum \infty \frac{2 n}{7 n - 6}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{7 n - 6}

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 - 6}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty n + 3

n = 0 \sum \infty \frac{7}{5 n + 4}

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

n = 0 \sum \infty \frac{3}{n !}

n = 0 \sum \infty 2^{n + 1}

What is the sum from n=1 to infinity of (2n)/(7n-6) ?
The sum from n=1 to infinity of (2n)/(7n-6) is diverges