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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= 7 diverges

= diverges

n = 2 \sum \infty (\frac{1}{8})^{n + 4}

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

n = 1 \sum \infty 6 - 5 (0.7)^{n}

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

n = 0 \sum \infty (sin (3))^{n}

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