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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= \frac{1}{2} diverges

= diverges

n = 0 \sum \infty \frac{1}{4} (\frac{3}{4})^{n}

n = 1 \sum \infty 12 sin (\frac{1}{n})

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

n = 0 \sum \infty n^{- \frac{3}{2}}

n = 3 \sum \infty \frac{1}{( n - 2 ) ^{4}}

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