What is the sum from n=1 to infinity of (5n)/(6n+1) ?

The sum from n=1 to infinity of (5n)/(6n+1) is diverges

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sum from n=1 to infinity of (5n)/(6n+1)

n = 1 \sum \infty \frac{5 n}{6 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{5 n}{6 n + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{n}{6 n + 1}

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

n = 1 \sum \infty \frac{1}{6}

n = 1 \sum \infty \frac{1}{n ( n - 4 )}

n = 1 \sum \infty \frac{n}{e ^{n^{4}}}

n = 1 \sum \infty 4 (0.5)^{n - 1}

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

What is the sum from n=1 to infinity of (5n)/(6n+1) ?
The sum from n=1 to infinity of (5n)/(6n+1) is diverges