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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty 2 (\frac{1}{5})^{n + 1}

k = 1 \sum \infty \frac{1}{k ( k + 2 )}

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

n = 0 \sum \infty (0.03)^{n}

k = 1 \sum \infty (- \frac{6}{7})^{k}

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