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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty ln (n + 1)

n = 0 \sum \infty \frac{1}{9 + n}

n = 0 \sum \infty \frac{n}{( n + 1 ) !}

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

n = 1 \sum \infty \frac{n + 1}{2 n + 3}

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