What is the sum from n=1 to infinity of n/(n^2-2n+5) ?

The sum from n=1 to infinity of n/(n^2-2n+5) is diverges

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sum from n=1 to infinity of n/(n^2-2n+5)

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty \frac{7}{1 + n ^{2}}

n = 0 \sum \infty 3

n = 1 \sum \infty \frac{2 n}{7 n - 9}

n = 1 \sum \infty \frac{2 n}{7 n - 5}

n = 1 \sum \infty \frac{1}{n ^{\frac{11}{2}}}

What is the sum from n=1 to infinity of n/(n^2-2n+5) ?
The sum from n=1 to infinity of n/(n^2-2n+5) is diverges