What is the sum from n=1 to infinity of 4-9(0.6)^n ?

The sum from n=1 to infinity of 4-9(0.6)^n is diverges

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sum from n=1 to infinity of 4-9(0.6)^n

n = 1 \sum \infty 4 - 9 (0.6)^{n}

diverges

Solution steps

n = 1 \sum \infty 4 - 9 \cdot 0. 6^{n}

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty \frac{6}{n ( n - 1 )}

n = 0 \sum \infty (e)^{- n}

n = 0 \sum \infty (\frac{4}{25})^{n}

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

k = 1 \sum \infty \frac{k}{500 k + 5}

What is the sum from n=1 to infinity of 4-9(0.6)^n ?
The sum from n=1 to infinity of 4-9(0.6)^n is diverges