What is the sum from n=1 to infinity of (5n)/(4n+7) ?

The sum from n=1 to infinity of (5n)/(4n+7) is diverges

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sum from n=1 to infinity of (5n)/(4n+7)

n = 1 \sum \infty \frac{5 n}{4 n + 7}

diverges

Solution steps

n = 1 \sum \infty \frac{5 n}{4 n + 7}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{n}{4 n + 7}

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

\frac{8}{( s ^{2} + 1 ) ( s ^{2} - 1 )}

\int \frac{a ^{2} - x ^{2}}{x ^{4}} d x

x \to 2 lim (x + 3)

y = ln (lo g_{e} (2 x))

n = 1 \sum \infty (x - 2)^{n}

What is the sum from n=1 to infinity of (5n)/(4n+7) ?
The sum from n=1 to infinity of (5n)/(4n+7) is diverges