What is the sum from n=0 to infinity of 1/(3n) ?

The sum from n=0 to infinity of 1/(3n) is diverges

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sum from n=0 to infinity of 1/(3n)

n = 0 \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

y = \frac{( 3 x - 6 )}{( x ^{2} + 4 )}

x^{2} + y^{3} = 12

\frac{d}{d x} (9 e^{x} + 5)

\frac{16}{( s + 4 ) ^{2}}

\frac{d y}{d x} - 5 y = 0

What is the sum from n=0 to infinity of 1/(3n) ?
The sum from n=0 to infinity of 1/(3n) is diverges