What is the sum from n=1 to infinity of (4n)/(7n-1) ?

The sum from n=1 to infinity of (4n)/(7n-1) is diverges

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sum from n=1 to infinity of (4n)/(7n-1)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{n}{7 n - 1}

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

\int \frac{1}{8 x x ^{2} - 64} d x

y^{^{'}} = \frac{ln ( 2 )}{7} y - 20000, y (0) = 200000

\frac{d}{d x} (x (x^{4} + 9)^{3})

\int ln^{2} (\frac{1}{2} x) d x

\frac{( 12 s + 8 )}{( s - 1 ) ^{5}}

What is the sum from n=1 to infinity of (4n)/(7n-1) ?
The sum from n=1 to infinity of (4n)/(7n-1) is diverges