What is the sum from n=1 to infinity of (2n)/(7n-3) ?

The sum from n=1 to infinity of (2n)/(7n-3) is diverges

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sum from n=1 to infinity of (2n)/(7n-3)

n = 1 \sum \infty \frac{2 n}{7 n - 3}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{7 n - 3}

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 - 3}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty \frac{1}{n ^{0}}

n = 2 \sum \infty \frac{7}{n ( n + 1 )}

n = 0 \sum \infty 0.2 5^{n}

n = 1 \sum \infty sin^{n} (3)

n = 1 \sum \infty sin^{n} (9)

What is the sum from n=1 to infinity of (2n)/(7n-3) ?
The sum from n=1 to infinity of (2n)/(7n-3) is diverges