What is the sum from n=1 to infinity}(4^{1/3 of)^n ?

The sum from n=1 to infinity}(4^{1/3 of)^n is diverges

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sum from n=1 to infinity}(4^{1/3 of)^n

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

diverges

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{6}{n !}

n = 1 \sum \infty (\frac{5}{4})^{n}

n = 1 \sum \infty \frac{e ^{\frac{6}{n}}}{n}

n = 1 \sum \infty \frac{2}{n ^{2} + 4}

n = 1 \sum \infty \frac{1}{n ( n + 4 )}

What is the sum from n=1 to infinity}(4^{1/3 of)^n ?
The sum from n=1 to infinity}(4^{1/3 of)^n is diverges