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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty 6^{- n}

k = 2 \sum \infty (- 0.45)^{k}

k = 1 \sum \infty \frac{4}{( 2 k - 1 ) ( 2 k + 1 )}

n = 0 \sum \infty ln (\frac{n}{3 n + 1})

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

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