What is the sum from n=1 to infinity of 1/(3n^4+12n) ?

The sum from n=1 to infinity of 1/(3n^4+12n) is converges

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sum from n=1 to infinity of 1/(3n^4+12n)

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

converges

Solution steps

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

Apply Series Comparison Test:

converges

= converges

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

n = 0 \sum \infty (\frac{5}{6})^{n - 1}

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

n = 0 \sum \infty \frac{6}{7 ^{n}}

x = 1 \sum \infty \frac{2 ^{x}}{5 ^{x}}

What is the sum from n=1 to infinity of 1/(3n^4+12n) ?
The sum from n=1 to infinity of 1/(3n^4+12n) is converges