What is the sum from n=1 to infinity of 1/(4n) ?

The sum from n=1 to infinity of 1/(4n) is diverges

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sum from n=1 to infinity of 1/(4n)

n = 1 \sum \infty \frac{1}{4 n}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{4 n}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{4} diverges

= diverges

\frac{d}{d x} (x^{3} - \frac{3 x ^{2}}{2})

\int \frac{1}{20} x d x

y^{^{'}} = 2 y^{2} sin (x)

(t - 2)^{2}

x \to c lim (\frac{x - c}{2 x - 2 c})

What is the sum from n=1 to infinity of 1/(4n) ?
The sum from n=1 to infinity of 1/(4n) is diverges