What is the sum from n=1 to infinity of 3n ?

The sum from n=1 to infinity of 3n is diverges

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sum from n=1 to infinity of 3n

n = 1 \sum \infty 3 n

diverges

Solution steps

n = 1 \sum \infty 3 n

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\frac{d}{d x} (300 x (1 - x))

\int x csc (x) cot (x) d x

\int_{0}^{8} x + 1 d x

\int sec (2 x + 1) tan (2 x + 1) d x

x \to 1 + lim (ln^{x - 1} (x))

What is the sum from n=1 to infinity of 3n ?
The sum from n=1 to infinity of 3n is diverges