What is the sum from n=1 to infinity of (n+1)/(3-2n) ?

The sum from n=1 to infinity of (n+1)/(3-2n) is diverges

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sum from n=1 to infinity of (n+1)/(3-2n)

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{1 1 ^{n}}{n !}

n = 0 \sum \infty \frac{i}{n ^{2}}

k = 1 \sum \infty (cos (14))^{k}

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

n = 0 \sum \infty n e^{- 2 n^{2}}

What is the sum from n=1 to infinity of (n+1)/(3-2n) ?
The sum from n=1 to infinity of (n+1)/(3-2n) is diverges