What is the sum from n=1 to infinity of (9n)/(6n+1) ?

The sum from n=1 to infinity of (9n)/(6n+1) is diverges

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sum from n=1 to infinity of (9n)/(6n+1)

n = 1 \sum \infty \frac{9 n}{6 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{9 n}{6 n + 1}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n}{6 n + 1}

Apply Series Divergence Test:

diverges

= 9 diverges

= diverges

0 = 1 \sum \infty \frac{ln ( 0 )}{0}

n = 1 \sum \infty \frac{n ^{4}}{7 ^{n}}

n = 0 \sum \infty (\frac{5}{4})^{n - 1}

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

k = 5 \sum \infty \frac{1}{( k - 4 ) ^{5}}

What is the sum from n=1 to infinity of (9n)/(6n+1) ?
The sum from n=1 to infinity of (9n)/(6n+1) is diverges