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

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

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

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

converges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{n ^{n}} = n^{- n}

= n = 2 \sum \infty n^{- n}

Apply Series Root Test:

converges

= converges

n \to \infty lim (n^{1})

\frac{\partial}{\partial x} (\frac{1}{( x ^{2} + y ^{2} + 1 )})

f (x) = x^{2} - 5 x + 5, (4, 1)

\int \frac{d}{1} (e^{x} cos (x)) d x

\frac{d}{d x} (\frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1)

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