Does this series converge or diverge : $\sum_{n=1}^\infty (\sqrt[n]n -1)$
Does summation nth root of n -1 converge?
2
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sequences-and-series
convergence
2 Answers
1
The series diverges by the limit comparison test since
$$\lim_{n \to \infty}\frac{n^{1/n} - 1}{1/n} = \lim_{n \to \infty} n(n^{1/n} - 1) =\infty.$$
We have $n^{1/n} > 1$ and $n^{1/n} = 1 + a_n$ with $a_n > 0$.
Hence,
$$n = (1 + a_n)^n\\ \implies \ln n = n \ln(1 + a_n) < na_n \\ \implies n(n^{1/n} -1 ) = n a_n > \ln n \to \infty.$$
7
Hint. One may observe that, as $n \to \infty$, $$ \sqrt[n]n -1=e^{\frac{\ln n}n}-1\ge\frac{\ln n}n\ge\frac{1}n. $$