What is the sum from n=1 to infinity of (8n)/(8n+1) ?

The sum from n=1 to infinity of (8n)/(8n+1) is diverges

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sum from n=1 to infinity of (8n)/(8n+1)

n = 1 \sum \infty \frac{8 n}{8 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{8 n}{8 n + 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{n}{8 n + 1}

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

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

n = 1 \sum \infty \frac{9}{1 + n ^{2}}

n = 1 \sum \infty \frac{1}{n ^{2} - 4}

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

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

What is the sum from n=1 to infinity of (8n)/(8n+1) ?
The sum from n=1 to infinity of (8n)/(8n+1) is diverges