What is the sum from n=1 to infinity of (9n)/(10) ?

The sum from n=1 to infinity of (9n)/(10) is diverges

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sum from n=1 to infinity of (9n)/(10)

n = 1 \sum \infty \frac{9 n}{10}

diverges

Solution steps

n = 1 \sum \infty \frac{9 n}{10}

Apply the constant multiplication rule:

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

= \frac{9}{10} \cdot n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= \frac{9}{10} diverges

= diverges

(- 10, - 3), (4, - 3)

\int e^{9 x - 14} d x

g (x) = e^{3}

\frac{\partial}{\partial x} (3 e^{x} ln (7))

f (x) = 4 t^{- \frac{3}{8}}

What is the sum from n=1 to infinity of (9n)/(10) ?
The sum from n=1 to infinity of (9n)/(10) is diverges