What is the sum from n=1 to infinity of n(9/8)^n ?

The sum from n=1 to infinity of n(9/8)^n is diverges

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sum from n=1 to infinity of n(9/8)^n

n = 1 \sum \infty n (\frac{9}{8})^{n}

diverges

Solution steps

n = 1 \sum \infty n (\frac{9}{8})^{n}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty (\frac{5}{12})^{n}

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

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

n = 5 \sum \infty \frac{1}{( n - 4 ) ^{3}}

k = 1 \sum \infty \frac{1}{5 ^{k} - 2 ^{k}}

What is the sum from n=1 to infinity of n(9/8)^n ?
The sum from n=1 to infinity of n(9/8)^n is diverges