What is the sum from n=1 to infinity of 1/(2(n+1)) ?

The sum from n=1 to infinity of 1/(2(n+1)) is diverges

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sum from n=1 to infinity of 1/(2(n+1))

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Limit Comparison Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 0 \sum \infty 0.99 9^{n}

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

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

x = 0 \sum \infty \frac{1}{x}

n = 1 \sum \infty (0.8)^{n - 1}

What is the sum from n=1 to infinity of 1/(2(n+1)) ?
The sum from n=1 to infinity of 1/(2(n+1)) is diverges