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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty sin^{n} (100)

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

n = 0 \sum \infty (0.6)^{n}

n = 1 \sum \infty \frac{9}{e ^{n}}

n = 1 \sum \infty \frac{8 n}{5 n + 1}

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