A college student's dream

Calculus Level 3

n = 1 1 n n \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^n}

Evaluate the summation above to three decimal places.


The answer is 1.29129.

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2 solutions

Abhishek Sinha
Apr 2, 2015

Note that the successive the terms in the series rapidly converges to zero. As an estimate, we have n = 6 1 n n x = 5 1 x 6 d x = 1 15625 = 0.000064 \sum_{n=6}^{\infty} \frac{1}{n^n} \leq \int_{x=5}^{\infty}\frac{1}{x^6} dx = \frac{1}{15625}= 0.000064 Hence to obtain a result correct upto the third term of the decimal, it suffices to compute the partial sum of the series only up to the first 5 terms, which evaluates to 1.291.

Moderator note:

Correct. Note that this summation is derived from Sophomore's dream thus the slightly subtle title to this problem.

Curtis Clement
Apr 12, 2015

It is clear that the infinite sum converges because : n = 1 1 n n < n = 1 1 n 2 = π 2 6 1.64 \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^n} < \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.64 The first 7 terms produces 1.291 to 3 significant figures

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