What is the sum from n=2 to infinity}((3^{3n of))/(2^{2n-2)} ?

The sum from n=2 to infinity}((3^{3n of))/(2^{2n-2)} is diverges

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sum from n=2 to infinity}((3^{3n of))/(2^{2n-2)}

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

diverges

Solution steps

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

Simplify

2^{2 n - 2} : 2^{2 n} \cdot 2^{- 2}

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

Apply the constant multiplication rule:

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

= \frac{1}{2 ^{- 2}} \cdot n = 2 \sum \infty \frac{3 ^{3 n}}{2 ^{2 n}}

\frac{1}{2 ^{- 2}} = 4

= 4 \cdot n = 2 \sum \infty \frac{3 ^{3 n}}{2 ^{2 n}}

Simplify

\frac{3 ^{3 n}}{2 ^{2 n}} : (\frac{27}{4})^{n}

= 4 \cdot n = 2 \sum \infty (\frac{27}{4})^{n}

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

= 4 (n = 0 \sum \infty (\frac{27}{4})^{n} - (\frac{27}{4})^{0} - (\frac{27}{4})^{1})

n = 0 \sum \infty (\frac{27}{4})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 10 lim (\frac{x}{x ^{2} - 100})

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\frac{d}{d x} (\frac{1 - cos ( x )}{x})

\int \frac{2 x - 1}{2 x + 3} d x

\frac{d}{d x} (ln (x + 2 y + 3 z))

What is the sum from n=2 to infinity}((3^{3n of))/(2^{2n-2)} ?
The sum from n=2 to infinity}((3^{3n of))/(2^{2n-2)} is diverges