What is the sum from n=4 to infinity of n/(n^2-5n+6) ?

The sum from n=4 to infinity of n/(n^2-5n+6) is diverges

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sum from n=4 to infinity of n/(n^2-5n+6)

n = 4 \sum \infty \frac{n}{n ^{2} - 5 n + 6}

diverges

Solution steps

n = 4 \sum \infty \frac{n}{n ^{2} - 5 n + 6}

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty 5^{- n} n!

n = 0 \sum \infty \frac{n}{9 n ^{2} + 1}

k = 0 \sum \infty (\frac{3}{5})^{k}

n = 1 \sum \infty 5 (\frac{1}{π})^{n}

n = 1 \sum \infty \frac{6}{n + 5}

What is the sum from n=4 to infinity of n/(n^2-5n+6) ?
The sum from n=4 to infinity of n/(n^2-5n+6) is diverges