What is the value of this series? $$ \sum_{k=1}^\infty \frac{1}{k^k} = 1 + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \cdots $$
I found the answer to be equal to $1.29$ by converting the sum to an integral.
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1@BAI, clearly, it's the latter. – Prasun Biswas Aug 27 '17 at 05:46
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6If it's latter, look up Sophomore's dream – BAI Aug 27 '17 at 05:47
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Actually you can get four exact decimal adding the first five terms $$1+\frac{1}{4}+\frac{1}{27}+\frac{1}{256}+\frac{1}{3125}\approx 1.2912$$ – Raffaele Aug 27 '17 at 11:29