# Convergence of $\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k^2}-1\right)$ and $\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k}-1\right)$ for $a>1$

I am reading a real analysis book which in the chapter about sequences and series asks to prove that, for $$a>1$$, $$\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k^2}-1\right)$$ is convergent and $$\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k}-1\right)$$ is divergent.

My proof.

Since for $$a>1$$ the sequence $$\{a^\frac{1}{k^2}\}$$ is positive and monotone decreasing and $$\int_{x=1}^{x=+\infty}(a^\frac{1}{x^2}-1)<+\infty$$ we have that $$\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k^2}-1\right)$$ is convergent by the integral test.

Alternatively, $$a^\frac{1}{k^2}-1=e^\frac{\ln(a)}{k^2}-1$$ and $$\lim\limits_{k\to+\infty}\frac{e^\frac{\ln(a)}{k^2}-1}{\frac{\ln(a)}{k^2}}=1$$ and since $$\sum\limits_{k=1}^{\infty}\frac{\ln(a)}{k^2}<+\infty$$ the claim follows by the limit comparison test.

Using the inequality $$e^x\geq 1+x\ \forall x \in\mathbb{R}$$ we also have that $$a^\frac{1}{k}-1=e^{\frac{\ln(a)}{k}}-1\geq\frac{\ln(a)}{k}$$ and since $$\sum\limits_{k=1}^{\infty}\frac{\ln(a)}{k}=+\infty$$ it follows that $$\sum\limits_{k=1}^{\infty}(a^\frac{1}{k}-1)=+\infty$$ by comparison test.

Now, while showing that $$\sum\limits_{k=1}^{\infty}(a^\frac{1}{k}-1)=+\infty$$ follows from two fairly elementary results, showing that $$\sum\limits_{k=1}^{\infty}\left(a^\frac{1}{k^2}-1\right)<+\infty$$ involves either the limit $$\lim\limits_{x\to 0}\frac{e^x-1}{x}=1$$ or more sophisticated machinery like improper integrals and the integral test, but both of these topics, in the real analysis book I am reading are explained only several chapters later, so I would like to know if there is a proof of this last result I have mentioned that doesn't use either limits of functions and/or integrals, thanks.

• Use $a^{\frac{1}{k^2}}-1\sim \log a\frac{1}{k^2}$ and $a^{\frac{1}{k}}-1\sim\log a\frac{1}{k}$ Sep 26, 2023 at 11:11
• $\lim_{x\to 0}\frac{e^x-1}{x}=1$ is the derivative of the exponential function at $x=0$, is that really not available at that point in the book? Sep 26, 2023 at 11:19
• @MartinR Yes, limits of functions are discussed only in the following section of the book, several chapters later. The book does give an hint, though, namely to use the identity $(a^\frac{1}{k}-1)\cdot\sum\limits_{h=0}^{k-1}a^\frac{h}{k}=a-1.$ Sep 26, 2023 at 11:22

From the lower bound for the exponential function $$e^x \ge 1+x$$ one can derive an upper bound as well: For $$x < 1$$ is $$e^x = \frac{1}{e^{-x}} \le \frac{1}{1-x} \, .$$ That can be used to prove the convergence of $$\sum_{k=1}^{\infty}(a^{1/k^2}-1)$$ in a similar way as you proved the divergence of $$\sum_{k=1}^{\infty}(a^{1/k}-1)$$:
For all sufficiently large $$k$$ is $$\ln(a)/k^2 \le 1/2$$, so that $$a^{1/k^2}-1 = e^{\ln(a)/k^2} - 1 \le \frac{1}{1-\ln(a)/k^2} - 1 \\ = \frac{\ln(a)/k^2}{1-\ln(a)/k^2} \le \frac{2}{k^2} \ln(a)$$ and the convergence follows by comparison with $$\sum_{k=1}^\infty 1/k^2$$.
Another solution, using the hint $$(a^{1/k}-1)\cdot\sum_{h=0}^{k-1}a^{h/k}=a-1$$:
$$a^{1/k}-1 = \frac{a-1}{\sum_{h=0}^{k-1}a^{h/k}} \ge \frac{a-1}{ka}$$ since $$a^{h/k} \le a$$, and $$a^{1/k^2}-1 = \frac{a-1}{\sum_{h=0}^{k^2-1}a^{h/k^2}} \le \frac{a-1}{k^2}$$ since $$a^{h/k^2} > 1$$.