What is the sum from n=1 to infinity of (2n)/(7n-5) ?

The sum from n=1 to infinity of (2n)/(7n-5) is diverges

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sum from n=1 to infinity of (2n)/(7n-5)

n = 1 \sum \infty \frac{2 n}{7 n - 5}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{7 n - 5}

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 - 5}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{1}{n ^{\frac{11}{2}}}

n = 1 \sum \infty \frac{n}{( - 2 ) ^{n}}

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

n = 1 \sum \infty n - 2

n = 3 \sum \infty \frac{6}{( n ^{2} - 1 )}

What is the sum from n=1 to infinity of (2n)/(7n-5) ?
The sum from n=1 to infinity of (2n)/(7n-5) is diverges