Competition problem (unknown source)

Question

For what positive $x$ does the series $$(x-1)+( \sqrt[2]{x}-1)+ ( \sqrt[3]{x}-1)+ … + ( \sqrt[n]{x}-1) + …$$ converge?

Answer 1

Clearly, the series converges for $x = 1$. Now we will show that it does not converge for $x \neq1$. Let us divide through by $x - 1$ and compare to the harmonic series. By the mean value theorem applied to $f(t) = t^{\frac{1}{n}}$, for each $n$ there exists $c_n$ between $x$ and $1$ such that $$\frac{\sqrt[n]{x}-1}{x-1} = \frac{1}{n}c^{\frac{1}{n}-1}.$$ Then, it follows that $$\frac{\sqrt[n]{x}-1}{x-1} > \frac{1}{n} (\text{max} (1,x))^ {\frac{1}{n}-1} > \frac{1}{n}(\text{max}(1,x))^ {-1}.$$ Summing, we find $$\sum_{n=1}^{\infty}\frac{\sqrt[n]{x}-1}{x-1} \geq (\text{max}(1,x))^ {-1} \sum_{n=1}^{\infty} \frac{1}{n} = \infty, $$ which proves that the series diverges.

References:

1. G.T. Gilbert, M.I. Krusemeyer, and L.C. Larson, The Wohascum County Problem Book, MAA, 1996.
2. R. Gelca and T. Andreescu, Putnam and Beyond, Springer, 2007.