What is the sum from n=0 to infinity of (n+1)/(2n) ?

The sum from n=0 to infinity of (n+1)/(2n) is diverges

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sum from n=0 to infinity of (n+1)/(2n)

n = 0 \sum \infty \frac{n + 1}{2 n}

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

\frac{n + 1}{n} = (\frac{n}{n} + \frac{1}{n})

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

Simplify

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

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

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

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

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

n = 1 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

n = 1 \sum \infty e^{- n} + e^{n}

What is the sum from n=0 to infinity of (n+1)/(2n) ?
The sum from n=0 to infinity of (n+1)/(2n) is diverges