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.

  • 1
    $\begingroup$ Use $a^{\frac{1}{k^2}}-1\sim \log a\frac{1}{k^2}$ and $a^{\frac{1}{k}}-1\sim\log a\frac{1}{k}$ $\endgroup$
    – Ychen
    Sep 26, 2023 at 11:11
  • $\begingroup$ $\lim_{x\to 0}\frac{e^x-1}{x}=1$ is the derivative of the exponential function at $x=0$, is that really not available at that point in the book? $\endgroup$
    – Martin R
    Sep 26, 2023 at 11:19
  • 1
    $\begingroup$ @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 $(a^\frac{1}{k}-1)\cdot\sum\limits_{h=0}^{k-1}a^\frac{h}{k}=a-1.$ $\endgroup$
    – lorenzo
    Sep 26, 2023 at 11:22

1 Answer 1


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 $a^{h/k^2} > 1$.


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