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

The sum from n=1 to infinity}(n^{1/n of-1)^n is converges

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

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 0 \sum \infty \frac{n}{1 + n ^{4}}

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

n = 3 \sum \infty n^{- 0.9999}

n = 2 \sum \infty (- 0.85)^{n}

n = 0 \sum \infty \frac{1}{( 3 n + 1 ) ^{3}}

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