What is the sum from n=1 to infinity of 2(-4/3)^n ?

The sum from n=1 to infinity of 2(-4/3)^n is diverges

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sum from n=1 to infinity of 2(-4/3)^n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

diverges

= 2 diverges

= diverges

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

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

n = 1 \sum \infty \frac{1}{n ^{2} + 30}

n = 0 \sum \infty \frac{1}{( 1 - n ) ^{2}}

k = 1 \sum \infty \frac{ln ( k )}{k}

What is the sum from n=1 to infinity of 2(-4/3)^n ?
The sum from n=1 to infinity of 2(-4/3)^n is diverges