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

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

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

n = 1 \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 \frac{1}{8 n}

n = 1 \sum \infty \frac{5 n}{n ^{2} + 9}

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

n = 1 \sum \infty n^{2} (\frac{5}{8})^{n}

n = 1 \sum \infty \frac{5 ^{n}}{2 n - 1}

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