What is the sum from n=2 to infinity of 1/(2n) ?

The sum from n=2 to infinity of 1/(2n) is diverges

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sum from n=2 to infinity of 1/(2n)

n = 2 \sum \infty \frac{1}{2 n}

diverges

Solution steps

n = 2 \sum \infty \frac{1}{2 n}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 2 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty 12 sin^{n} (7 0^{\circ})

n = 0 \sum \infty \frac{n}{5 n ^{2} + 1}

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

n = 0 \sum \infty (\frac{4 π}{e})^{n}

k = 0 \sum \infty \frac{k}{3 k + 1}

What is the sum from n=2 to infinity of 1/(2n) ?
The sum from n=2 to infinity of 1/(2n) is diverges