What is the sum from n=1 to infinity of 3n^{-1} ?

The sum from n=1 to infinity of 3n^{-1} is diverges

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sum from n=1 to infinity of 3n^{-1}

n = 1 \sum \infty 3 n^{- 1}

diverges

Solution steps

n = 1 \sum \infty 3 n^{- 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty n^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= 3 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 3 diverges

= diverges

k = 2 \sum \infty (- 0.75)^{k}

n = 1 \sum \infty n^{- 0.3}

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

x = 0 \sum \infty \frac{x}{2 ^{x}}

x = 0 \sum \infty \frac{1}{x !}

What is the sum from n=1 to infinity of 3n^{-1} ?
The sum from n=1 to infinity of 3n^{-1} is diverges