Is there any lower bound on a convergent, non-negative series?

For example, if $\sum_{n=1}^\infty a_n$ is convergent, and $a_n \geq 0$ $\forall n$ in the bounds $\{1, 2, 3, 4, ...\}$, then can it be said that $a_n \leq \frac{1}{n^{1+\epsilon}}$, for arbitrarily small $\epsilon > 0$.

I am seeing if this holds generally, since $\sum_{n=1}^\infty \frac{1}{n}$ diverges (harmonic series), and $\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$ converges $\forall \epsilon > 0$.

Therefore, is it the case that $\frac{1}{n^{1+\epsilon}} \geq a_n$, for some $\epsilon > 0$, given that the sum $\sum_{n=1}^\infty a_n$ converges? I.e., in general, is it the case that for a non-negative series that converges, that some power greater than 1 for the denominator can be found such that $a_n \leq \frac{1}{n^{1+\epsilon}}$?

  • $\begingroup$ This should occur only for the $a_n$'s such that $a_n= o\left(\frac{1}{n^{1+\varepsilon}}\right)$. $\endgroup$ Nov 22, 2020 at 22:06
  • $\begingroup$ I was wondering about this, since $\sin{\frac{\ln{k}}{k^2}}$ is still bounded by this, as well as $(\sqrt[k]{k}-1)^k$. Taking the sum of either of these series from $n = 1 \to \infty$ converges, and the terms are both non-negative and bounded of the form $\frac{1}{n^{1+\epsilon}}$ $\endgroup$
    – qxzsilver
    Nov 22, 2020 at 22:25
  • $\begingroup$ I think you may have meant upper bound here? I posted an answer on lower bound but it was deleted, so maybe I am just not understanding. My thinking is that surely a <= b is not a lower bound and 0 (of course) is the (only) lower bound for all positive series. $\endgroup$
    – skaak
    Nov 23, 2020 at 14:07

1 Answer 1


The answer is no, the simplest counterexample is

$$\sum_{n=1}^\infty \frac{1}{n (1+\ln(n)^{1+\delta})}$$ for any $\delta >0$.

This series is convergent, but it is not bounded by any $\frac{1}{n^{1+\epsilon}}$.

And you can take this one step further $\sum_{n=1}^\infty \frac{1}{n (1+\ln(n))(1+\ln(\ln(n)))^{1+\delta})}$ and so on.

  • $\begingroup$ Thanks for the comment. Interesting choice of the series - is there some unique algebraic or analytic property to these family of series (or do these types of series have a name)? $\endgroup$
    – qxzsilver
    Nov 22, 2020 at 22:29
  • $\begingroup$ @qxzsilver Not really. The basic idea is that if $f(n)$ is a decreasing positive function such that $\sum f(n)$ is convergent, then $sum \farc{1}{n} f(\ln(n))$ converges by a double application of the integral test, and is often slower than $f(n)$. I just used this approach on your series, I only added the 1+ to make the series start at 1 instead of 2, but that is not important. $\endgroup$
    – N. S.
    Nov 22, 2020 at 23:57

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