What is the sum from n=1 to infinity of (n^2)/(n^4) ?

The sum from n=1 to infinity of (n^2)/(n^4) is converges

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sum from n=1 to infinity of (n^2)/(n^4)

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

converges

Solution steps

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

Simplify

\frac{n ^{2}}{n ^{4}} : \frac{1}{n ^{2}}

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

Apply p-Series Test:

converges

= converges

n = 5 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

n = 0 \sum \infty \frac{5 n !}{2 ^{n}}

n = 0 \sum \infty \frac{5}{6 ^{n}}

n = 2 \sum \infty (\frac{3}{8})^{4 n}

n = 0 \sum \infty \frac{4}{( 1 - 4 n ) ^{2}}

What is the sum from n=1 to infinity of (n^2)/(n^4) ?
The sum from n=1 to infinity of (n^2)/(n^4) is converges