What is the sum from n=0 to infinity of 7/(5n+4) ?

The sum from n=0 to infinity of 7/(5n+4) is diverges

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sum from n=0 to infinity of 7/(5n+4)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{5 n + 4}

Apply Series Limit Comparison Test:

diverges

= 7 diverges

= diverges

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

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

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

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

n = 0 \sum \infty \frac{4 ^{n}}{2 ^{n}}

What is the sum from n=0 to infinity of 7/(5n+4) ?
The sum from n=0 to infinity of 7/(5n+4) is diverges