It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values.

We define the operator $F_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert^{-1}} \text{ for }\varepsilon>0.$$

Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)_n$. Intuitively, the slow decay of the harmonic series should imply that it converges slower than anything summable (for most of it).

Therefore, I ask: Is it true that for any positive summable sequence $x$

$$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(x)}{F_{\varepsilon}((1/n))} \le 1?$$

It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values.

We define the operator $F_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert^{-1}} \text{ for }\varepsilon>0.$$

Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)_n$. Intuitively, the slow decay of the harmonic series should imply that it converges slower than anything summable (for most of it).

Therefore, I ask: Is it true that for any positive summable sequence $x$

$$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(x)}{F_{\varepsilon}((1/n))} \le 1?$$

Please just let me know if you have any questions.

It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values.

We define the operator $F_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert^{-1}} \text{ for }\varepsilon>0.$$

Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)_n$. Intuitively, the slow decay of the harmonic series should imply that it converges faster than anything summable (for most of it).

Therefore, I ask: Is it true that for any positive summable sequence $x$

$$\liminf_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(x)}{F_{\varepsilon}((1/n))} \ge 1?$$

It is well-known that the harmonic series is not summable. In some sense this means that it takes a lot of rather large values.

We define the operator $F_{\varepsilon}: \ell^{\infty}(\mathbb N) \rightarrow [0,\infty]$ by $$F_{\varepsilon}(x) = \sum_{i=1}^{\infty} 2^{-\varepsilon \vert x_i \vert^{-1}} \text{ for }\varepsilon>0.$$

Now consider a positive summable sequence $x$ and the harmonic sequence $(1/n)_n$. Intuitively, the slow decay of the harmonic series should imply that it converges slower than anything summable (for most of it).

Therefore, I ask: Is it true that for any positive summable sequence $x$

$$\limsup_{\varepsilon \downarrow 0} \frac{F_{\varepsilon}(x)}{F_{\varepsilon}((1/n))} \le 1?$$