What is the sum from n=1 to infinity of 8n ?

The sum from n=1 to infinity of 8n is diverges

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sum from n=1 to infinity of 8n

n = 1 \sum \infty 8 n

diverges

Solution steps

n = 1 \sum \infty 8 n

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

f (x) = (\frac{x ^{2} + 3}{x ^{2} - 3})^{4}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 5}

\int \frac{cos ( 3 x ) cos ( 2 x )}{3} d x

\int_{0}^{\frac{π}{6}} 20 sec^{4} (x) d x

\int \frac{8 x + 6}{2 x ^{2} + 3 x - 2} d x

What is the sum from n=1 to infinity of 8n ?
The sum from n=1 to infinity of 8n is diverges