What is the sum from n=0 to infinity of (3n)/(1+n^4) ?

The sum from n=0 to infinity of (3n)/(1+n^4) is converges

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sum from n=0 to infinity of (3n)/(1+n^4)

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

converges

Solution steps

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 0 \sum \infty \frac{n}{1 + n ^{4}}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty 6 - 5 (0.7)^{n}

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

n = 0 \sum \infty (sin (3))^{n}

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

n = 0 \sum \infty \frac{15}{6 ^{n}}

What is the sum from n=0 to infinity of (3n)/(1+n^4) ?
The sum from n=0 to infinity of (3n)/(1+n^4) is converges