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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 5 converges

= converges

\int_{- 1}^{3} 2 x^{2} + 1 d x

y^{^{'}} + \frac{1}{x} y = - \frac{1}{x} y^{2}

y = x^{3} - x

\frac{d}{d x} (\frac{2 ( x + 8 ) ( x - 6 )}{x ^{2} - 7 ^{2}})

\frac{d}{d x} ((\frac{2 x + 4}{3 x - 1})^{4})

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