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

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

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

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

converges

Solution steps

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

Simplify

\frac{1}{n} (\frac{1}{n})^{2} : \frac{1}{n ^{3}}

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

Apply p-Series Test:

converges

= converges

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

f (x) = 6 + 1 - x, at x = - 8

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

\frac{d}{d t} (e^{- 4 t})

x \frac{d y}{d x} + 3 x^{2} + 4 y = 0

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