What is the sum from n=1 to infinity of 1/(n^2-4n+5) ?

The sum from n=1 to infinity of 1/(n^2-4n+5) is converges

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sum from n=1 to infinity of 1/(n^2-4n+5)

n = 1 \sum \infty \frac{1}{n ^{2} - 4 n + 5}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} - 4 n + 5}

Apply Series Integral Test:

converges

= converges

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

n = 44 \sum \infty \frac{5}{n ^{2} + 2 n}

n = 1 \sum \infty 9 (\frac{4}{5})^{n}

n = 0 \sum \infty 2 (\frac{1}{4})^{n}

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

What is the sum from n=1 to infinity of 1/(n^2-4n+5) ?
The sum from n=1 to infinity of 1/(n^2-4n+5) is converges