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

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

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sum from n=1 to infinity of 2/n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty 3^{n} 4^{- n + 1}

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

n = 1 \sum \infty \frac{1}{1000 n + 1}

n = 5 \sum \infty \frac{8}{n ^{2} - 1}

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

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