Determine whether or not the series $\sum_{k=1}^{\infty}(\sqrt[k]{a} - 1)$ converges? Let $a > 1$ and determine whether or not the series $\sum_{k=1}^{\infty}(\sqrt[k]{a} - 1)$ converges?
I feel like I should use the Limit Comparison Test or the Comparison Test but I am having a difficult time finding a series i could use for that?
 A: $\begin{array}\\
\sum_{k=1}^{n}(\sqrt[k]{a} - 1)
&=\sum_{k=1}^{n}(e^{\ln(a)/k} - 1)\\
&=\sum_{k=1}^{n}((1+\ln(a)/k+O(1/k^2)) - 1)\\
&=\sum_{k=1}^{n}(\ln(a)/k+O(1/k^2))\\
&=\ln(a)\ln(n)+O(1)\\
\end{array}
$
and this diverges as
$n \to \infty$.
Note that
$\sum_{k=1}^{n}(\sqrt[k]{a} - 1-\frac{\ln(a)}{k})
$
converges.
A: Hints
Be sure you can supply details in the following:
$$a-1=(a^{1/k}-1)(a^{\frac{k-1}k}+a^{\frac{k-2}k}+\ldots+a^{1/k}+1)\implies$$
$$\sqrt[k]a-1=\frac{a-1}{a^{\frac{k-1}k}+a^{\frac{k-2}k}+\ldots+a^{1/k}+1}\stackrel{\text{since}\;a>1}\ge\frac{a-1}{ka^{\frac{k-1}k}}=$$
and using the limit comparison theorem:
$$\lim_{k\to\infty}\frac{\frac1{ka^{1-\frac1k}}}{\frac1k}=\lim_{k\to\infty}\frac{\sqrt[k]a}{a}=\frac1a\neq0$$
we get our series doesn't converge
A: The series diverges by comparison with the harmonic series. Note that
$$\frac{\log a}{k} = \log \sqrt[k]{a} = \log (1 + (\sqrt[k]{a}-1)) \leqslant \sqrt[k]{a} - 1,$$
and
$$\sum_{k=1}^n(\sqrt[k]{a} -1) \geqslant \log a \sum_{k=1}^n\frac{1}{k}$$
