What is the sum from n=1 to infinity of (8n)/(5n+1) ?

The sum from n=1 to infinity of (8n)/(5n+1) is diverges

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sum from n=1 to infinity of (8n)/(5n+1)

n = 1 \sum \infty \frac{8 n}{5 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{8 n}{5 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}{5 n + 1}

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

k = 1 \sum \infty 2 (- \frac{1}{5})^{2 k}

n = 1 \sum \infty n^{- \frac{5}{2}}

k = 1 \sum \infty sin (\frac{1}{k})

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

n = 1 \sum \infty cos (\frac{11}{n})

What is the sum from n=1 to infinity of (8n)/(5n+1) ?
The sum from n=1 to infinity of (8n)/(5n+1) is diverges