What is the sum from n=1 to infinity of n^{-1-1/n} ?

The sum from n=1 to infinity of n^{-1-1/n} is diverges

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sum from n=1 to infinity of n^{-1-1/n}

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

diverges

Solution steps

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

Apply exponent rule:

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

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

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

Simplify

n^{1 + \frac{1}{n}} : n n^{\frac{1}{n}}

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

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty - \frac{1}{1 5 ^{n}}

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

n = 0 \sum \infty e^{- 21 n}

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

n = 0 \sum \infty \frac{1}{n ^{2} + 9}

What is the sum from n=1 to infinity of n^{-1-1/n} ?
The sum from n=1 to infinity of n^{-1-1/n} is diverges