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

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

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sum from n=1 to infinity of 1/(3n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{3} diverges

= diverges

x \to - 5 + lim (8)

5 x^{3} - 4 x^{2} + 7 x - 8, at x = 1

\int_{0}^{1} (4 x + 2) e^{2 x} d x

\int 19 tan^{2} (x se) c^{3} x d x

f (x) = \frac{3 x}{x ^{2} - 1}

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