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

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

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

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= diverges

n = 2 \sum \infty \frac{1}{n ^{\frac{5}{3}}}

n = 2 \sum \infty (\frac{2}{7})^{n}

n = 0 \sum \infty n (n - 6)

n = 1 \sum \infty 6 (\frac{1}{8})^{n - 1}

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

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