What is the sum from n=1 to infinity of (9/4)^n ?

The sum from n=1 to infinity of (9/4)^n is diverges

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sum from n=1 to infinity of (9/4)^n

n = 1 \sum \infty (\frac{9}{4})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{9}{4})^{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 (\frac{9}{4})^{n} - (\frac{9}{4})^{0}

n = 0 \sum \infty (\frac{9}{4})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 2 \sum \infty \frac{1}{ln ( n ^{n} )}

n = 0 \sum \infty 2^{n} 3^{n - 2}

n = 5 \sum \infty \frac{ln ( n )}{n}

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

n = 3 \sum \infty 4 (\frac{1}{2})^{n + 2}

What is the sum from n=1 to infinity of (9/4)^n ?
The sum from n=1 to infinity of (9/4)^n is diverges