What is the sum from n=1 to infinity of (2n)/(5n-9) ?

The sum from n=1 to infinity of (2n)/(5n-9) is diverges

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sum from n=1 to infinity of (2n)/(5n-9)

n = 1 \sum \infty \frac{2 n}{5 n - 9}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{5 n - 9}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{5 n - 9}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

\int x^{4} - 6 x^{2} - 2 x + 4 d x

\int \frac{2 x ^{2} + 3}{x ^{3} - 2 x ^{2} + x} d x

\frac{1}{( x ^{2} - 2 x - 5 ) ^{4}}

\int_{0}^{2} \frac{1}{2} y^{2} d y

sin (\frac{t}{t + 9})

What is the sum from n=1 to infinity of (2n)/(5n-9) ?
The sum from n=1 to infinity of (2n)/(5n-9) is diverges