Does we have $\int_{\sum_{i=1}^{\frac{p}{k}}\frac{1}{i}}^{\sum_{i=1}^{p}\frac{1}{i}}e^{-\left(\frac{x}{p}\right)^{p}}dx\to \ln(k)$ as $p\to \infty$?

Question

Problem/Conjecture :

Let $p,k,n\in(3,\infty)$ be positive integers such that $p=kn$ then it seems we have as $p\to \infty$ :

$$\int_{\sum_{i=1}^{\frac{p}{k}}\frac{1}{i}}^{\sum_{i=1}^{p}\frac{1}{i}}e^{-\left(\frac{x}{p}\right)^{p}}dx\to \ln(k)$$

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I have tried to squeeze the integral with the Fatou-Lebesgue theorem but cannot proceed further here .

On the other hand the integral above implies the incomplete Gamma function but again I don't see how to get the logarithm .

It seems also to note this is the discrete case and I think the continued analogue is plausible too .

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Any ideas to (dis)prove it ?

Answer 1

With the usual notation $H_m:=\sum_{i=1}^m\frac1i$, and $0\leqslant 1-e^{-x}\leqslant x$ for $x\geqslant 0$, we have $$0\leqslant H_p-H_{p/k}-\int_{H_{p/k}}^{H_p}e^{-(x/p)^p}\,dx=\int_{H_{p/k}}^{H_p}\big(1-e^{-(x/p)^p}\big)dx\leqslant(H_p/p)^p(H_p-H_{p/k}),$$ and it remains to use $\lim\limits_{p\to\infty}(H_p-H_{p/k})=\ln k$ and $\lim\limits_{p\to\infty} H_p/p=0$.

Answer 2

Hint : we have with the same constraint :

$$\int_{\sum_{i=1}^{\frac{p}{k}}\frac{1}{i}}^{\sum_{i=1}^{p}\frac{1}{i}}e^{-\frac{x}{p}}dx<\int_{\sum_{i=1}^{\frac{p}{k}}\frac{1}{i}}^{\sum_{i=1}^{p}\frac{1}{i}}e^{-\left(\frac{x}{p}\right)^{p}}dx<\int_{\sum_{i=1}^{\frac{p}{k}}\frac{1}{i}}^{\sum_{i=1}^{p}\frac{1}{i}}e^{-\frac{x}{p^{p}}}dx$$

Now integrate and use the squeeze theorem as @Metamorphy .