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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply exponent rule:

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

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

= 2 \cdot n = 1 \sum \infty n^{- n}

Apply Series Root Test:

converges

= 2 converges

= converges

n = 2 \sum \infty (\frac{5}{6})^{3 n}

n = 1 \sum \infty \frac{1 8 ^{n}}{n !}

k = 0 \sum \infty (\frac{k}{k + 15})^{k}

k = 1 \sum \infty \frac{1}{k ^{6}}

k = 0 \sum \infty (0.8)^{k}

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