What is the sum from n=1 to infinity of (9n)/(7n+1) ?

The sum from n=1 to infinity of (9n)/(7n+1) is diverges

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sum from n=1 to infinity of (9n)/(7n+1)

n = 1 \sum \infty \frac{9 n}{7 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{9 n}{7 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}{7 n + 1}

Apply Series Divergence Test:

diverges

= 9 diverges

= diverges

y = x, y = 0, y = x - 2

x \to 3 + lim (\frac{2}{x - 3})

y^{^{'}} = y - 4 y^{2}

y^{^{'}} = \frac{x ^{2}}{y ( 1 + x ^{3} )}

\frac{d}{d x} (2^{c o t (5 x)})

What is the sum from n=1 to infinity of (9n)/(7n+1) ?
The sum from n=1 to infinity of (9n)/(7n+1) is diverges