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

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

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sum from n=1 to infinity of 3/n

n = 1 \sum \infty \frac{3}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{3}{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 3 diverges

= diverges

\int \frac{2}{t - t ^{3}} d t

\frac{d y}{d x} = \frac{9 x + 2 y}{9 x + 2 y + 2}

\frac{\partial}{\partial x} (x \frac{e ^{y}}{x + y})

x \to 0 lim (ln (x))

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