What is the sum from n=0 to infinity of 3+2/(n^4) ?

The sum from n=0 to infinity of 3+2/(n^4) is diverges

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sum from n=0 to infinity of 3+2/(n^4)

n = 0 \sum \infty 3 + \frac{2}{n ^{4}}

diverges

Solution steps

n = 1 \sum \infty 3 + \frac{2}{n ^{4}}

Simplify

3 + \frac{2}{n ^{4}} : \frac{3 n ^{4} + 2}{n ^{4}}

= n = 1 \sum \infty \frac{3 n ^{4} + 2}{n ^{4}}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{3}{n ^{9}}

n = 0 \sum \infty \frac{n ^{2}}{1 + 6 n}

n = 1 \sum \infty \frac{2}{n ^{9}}

n = 1 \sum \infty (0.75)^{n - 1}

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

What is the sum from n=0 to infinity of 3+2/(n^4) ?
The sum from n=0 to infinity of 3+2/(n^4) is diverges