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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 4 converges

= converges

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

n = 1 \sum \infty \frac{1}{n ^{- 6}}

n = 1 \sum \infty ln (\frac{n + 3}{n})

n = 1 \sum \infty - \frac{1}{1 3 ^{n}}

n = 0 \sum \infty \frac{1}{n} + \frac{1}{n + 1}

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