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

The sum from n=1 to infinity of 1/(3^{1/n)} is diverges

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sum from n=1 to infinity of 1/(3^{1/n)}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

x \to 3 lim (6 x - 7)

f (x) = x^{2} - 8, at x = 7

\frac{d}{d x} (8 x + 6)

f (x) = - 2 x + 5

\int \frac{1}{y ^{2} - 1} d y

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