What is the sum from n=0 to infinity of 1/(n^4+n^2) ?

The sum from n=0 to infinity of 1/(n^4+n^2) is converges

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sum from n=0 to infinity of 1/(n^4+n^2)

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

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

k = 2 \sum \infty (- 0.35)^{k}

n = 1 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

k = 5 \sum \infty 2 (- \frac{1}{4})^{k}

n = 4 \sum \infty 4 (6 - 2 n)

What is the sum from n=0 to infinity of 1/(n^4+n^2) ?
The sum from n=0 to infinity of 1/(n^4+n^2) is converges