# The lower bound of all positive convergent series

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 $$a_n \leq \frac{1}{n^{1+\epsilon}}$$?

• This should occur only for the $a_n$'s such that $a_n= o\left(\frac{1}{n^{1+\varepsilon}}\right)$. Nov 22, 2020 at 22:06
• 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 $\frac{1}{n^{1+\epsilon}}$ Nov 22, 2020 at 22:25
• 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. Nov 23, 2020 at 14:07

$$\sum_{n=1}^\infty \frac{1}{n (1+\ln(n)^{1+\delta})}$$ 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.
• @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. Nov 22, 2020 at 23:57