I'm given a sequence of moments $$ S_k=\int_{1}^{\infty}x^k \exp \left(\frac{-x}{\log(x)}\right)dx $$ and I'm told that this sequence is determined. However, I can't find a way to show this. I tried starting out by using Carleman's Condition, i.e. showing that $$\sum_{k=1}^{\infty} \frac{1}{S_{2k}^{1/(2k)}}=\infty.$$ Attempting this method has actually led me to believe the above sum might converge: so I wouldn't be able to use this condition to conclude "determinance". In any case, I wish to determine the convergence or divergence of the above series. My attempts to show this is by finding sequences such that $$A_k \leq S_k \leq B_k$$ and finding $f$ and $g$ so that $A_k=\int_{1}^{\infty}x^kfdx$ and $B_k=\int_{1}^{\infty}x^kgdx$- and $A_k$ and $B_k$ are convergent series for each $k$. I tried find functions that are either above or below $\frac{-x}{\log(x)}$ to get $f$ and $g$. What kind of $f$ and/or $g$ could I use here?

$\textbf{Edit (3)}:$ Attempt 2 on getting an upper bound is by taking the intervals $\big[ e^i, e^{i+1} \big]$ and having that $$\int_{1}^{\infty}x^k \exp \left(\frac{-x}{\log(x)}\right)dx \leq \sum_{i=0}^{\infty}\exp \left( (i+1)k-\frac{\exp \left( i+1 \right) }{i+1} \right) \left( e^{i+1} - e^{i} \right) \\ = \frac{e-1}{e}\sum_{i=0}^{\infty} \exp \left( (i+1)k-\frac{\exp \left( i+1 \right) }{i+1} +(i+1)\right)$$ Assuming $k \geq 5$, we have that $$\frac{e^{i+1}}{(i+1)(k+1)} - (i+1) \geq i$$ for every $i \geq k.$ So we can split the sum into $$\sum_{i=0}^{k-1} \exp \left( (i+1)k-\frac{\exp \left( i+1 \right) }{i+1} +(i+1)\right) + \sum_{i=k}^{\infty} \exp \left( -(k+1)\right) ^ {\frac{e^{i+1}}{(i+1)(k+1)} - (i+1)} \\ \leq k * max_{0 \leq i \leq k-1} \exp \left( (i+1)(k+1)-\frac{\exp \left( i+1 \right) }{i+1} \right) + \sum_{i=k}^{\infty} \exp \left( -(k+1)\right)^{i}$$ The right-most sum is geometric and less than 1 for any $k \geq 5$. One can calculate that the maximum is attained for $i$ such that $\frac{ie^{i+1}}{{(i+1)}^2}=k+1$. Hence, the exponent of the maximum becomes $$(i+1)(k+1)-\frac{(i+1)(k+1) }{i} = (i-\frac{1}{i})(k+1)$$ for our given $i$, and this is $O(k log(k))$. Hence, our sum overall is $O(k^2k^k)$, or subsequently $O(k^{2k})$ with plenty of room to spare. Hence by the Stielje's condition for moments on a half-line we have that $S_k$ is determined.

Does this look right?

  • $\begingroup$ You only need the upper bound. You need $S_k = \mathscr O(k^{2k})$. $\endgroup$ – Keith McClary Jun 12 at 4:55
  • $\begingroup$ @KeithMcClary I stated the Carlemann's condition incorrectly, so I believe your comment is insufficient now. Sorry! $\endgroup$ – Sean Nemetz Jun 12 at 5:30
  • $\begingroup$ Do you have the Stieltjes version for the half line? $\endgroup$ – Keith McClary Jun 12 at 5:46
  • $\begingroup$ @KeithMcClary Now I do :) . Now, I gotta figure out how to get that upper bound. $\endgroup$ – Sean Nemetz Jun 12 at 17:31
  • $\begingroup$ @KeithMcClary Sorry to ask. But do you have a hint to prove that upper bound? I feel like I'm overcomplicating it. $\endgroup$ – Sean Nemetz Jun 12 at 23:45

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