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

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

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

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 0 \sum \infty 2 \cdot 2^{n}

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

n = 0 \sum \infty \frac{11}{2 n - 1}

n = 1 \sum \infty (\frac{n}{10})^{n}

n = 1 \sum \infty \frac{1}{2} + \frac{4}{3 ^{n}}

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