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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 3 \sum \infty \frac{1}{n ( n + 2 )}

n = 0 \sum \infty \frac{1}{ln ( n ^{2} )}

n = 0 \sum \infty (\frac{i}{1 + i})^{n}

n = 1 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

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