Problem 282 (difficulty: 4/10)

Calculate \(\displaystyle \lim_{n\to \infty} \sqrt[n]{2^n-n}.\)

Solution:

\(\displaystyle 2=\sqrt[n]{2^n}>\sqrt[n]{2^n-n}>\sqrt[n]{2^n-2^{n-1}}=2\sqrt[n]{\frac12}, \)

for \(\displaystyle n\) big enough. The RHS tends to 2 by exercise 281, so the sandwich theorem implies the result.

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