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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

\frac{\partial}{\partial x} (x y (x - y)^{- 1})

x \to 99 lim (5)

x \to 0 + lim (csc (x) - \frac{1}{x})

t an g e n t f (x) = - \frac{1}{x}, at x = 4

x \to 4 lim (\frac{12 - 3 x}{x - 4})

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