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Given $$\lim_{x\to \infty} \frac 1x \sum_{n\le x} a_n = k,$$ I want to prove that $$\lim_{x\to \infty} \frac {1}{\log x} \sum_{n\le x} \frac {a_n}{n} = k.$$

I'm much more interested in learning the technique(s) necessary in order to prove this rather than a direct proof. Specifically, we have learned about asymptotic estimates and summation by parts, but I don't see how I can use those techniques to prove the problem statement. Thank you for any insight!

linearalgebrathrowaway
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2 Answers2

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I would say summation by parts is a useful general technique here. This would be very straightforward if $x$ is an integer.

In general, let $m = \lfloor x \rfloor$ and $S_m = \sum_{n=1}^m a_n$. Summing by parts we have

$$\tag{*}\sum_{n\leqslant x} \frac{a_n}{n} =\sum_{n=1}^m \frac{a_n}{n} = \frac{S_m}{m} + \sum_{n=1}^{m-1} S_n \left( \frac{1}{n} - \frac{1}{n+1}\right)= \frac{S_m}{m} + \sum_{n=1}^{m-1} \frac{S_n}{n(n+1)} $$

Dividing by $\log x$ and taking the limit as $x \to \infty$ we get the desired result.

Note that $\lim_{x \to \infty} m/x = 1$ since $x- 1 < m \leqslant x$, and

$$\lim_{m \to \infty} \frac{S_m}{m} = \lim_{x \to \infty} \frac{x}{m}\lim_{x \to \infty} \frac{1}{x}\sum_{n \leqslant x} a_n = k \\ \implies \lim_{x \to \infty} \frac{S_m}{m\log x} = 0$$

Applying the Stolz-Cesaro theorem, we can handle the limit of the second term on the RHS of (*).

$$\begin{align}\lim_{x \to \infty} \frac{1}{\log x}\sum_{n=1}^{m-1} \frac{S_n}{n(n+1)} &= \lim_{x \to \infty} \frac{\log m}{\log x} \lim_{m \to \infty}\frac{1}{\log m}\sum_{n=1}^{m-1} \frac{S_n}{n(n+1)}\\ &= \lim_{m \to \infty} \frac{ \sum_{n=1}^{m} \frac{S_n}{n(n+1)} -\sum_{n=1}^{m-1} \frac{S_n}{n(n+1)} }{\log(m+1) - \log m}\\ &= \lim_{m \to \infty} \frac{S_m/m}{(m+1)\log(1 +1/m)} \\ &= \frac{k}{\log e} = k\end{align}$$

Paramanand Singh
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RRL
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  • Very nice proof. +1 I have fixed a minor typo – Paramanand Singh Feb 28 '18 at 09:51
  • Thank you for the proof. Can you (or anyone else) explain how we can justify, in the last part of the proof, how $\lim_{m \to \infty}\frac{1}{\log m}\sum_{n=1}^{m-1} \frac{S_n}{n(n+1)}\ = \lim_{m \to \infty} \frac{ \sum_{n=1}^{m} \frac{S_n}{n(n+1)} -\sum_{n=1}^{m-1} \frac{S_n}{n(n+1)} }{\log(m+1) - \log m}$ Is it from the Stolz-Cesaro theorem? From the statements of the theorem I've seen, I don't see justification for $\lim_{n \to \infty}\frac{a_n}{b_n} = \lim_{n \to \infty}\frac{a_{n+1} - a_n}{b_{n+1} - b_n}$? – linearalgebrathrowaway Feb 28 '18 at 16:40
  • Specifically, I've seen if $\lim_{n \to \infty}\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \ell $, then $\lim_{n \to \infty}\frac{a_n}{b_n} = \ell $, but this is the reverse direction of how you seem to be applying the theorem. – linearalgebrathrowaway Feb 28 '18 at 17:12
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    @linearalgebrathrowaway: So $a_m = \sum_{n=1}^{m-1} \frac{S_n}{n(n+1)}$ and $b_m = \log m \uparrow \infty$. I'm proving $\lim \frac{a_m}{b_m} = k$ by showing that $\frac{a_{m+1} - a_m}{b_{m+1} - b_m} = \frac{\frac{S_m}{m(m+1)}}{\log(1+1/m)} \to k$. – RRL Feb 28 '18 at 17:22
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    I'm not using the converse (which is sometimes true in special situations). See this question – RRL Feb 28 '18 at 17:24
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Let $A(x)$ be defined as

$$ A(x)=\sum_{n\le x}a_n=kx+o(x) $$

then we have

$$ \begin{aligned} \sum_{n\le x}{a_n\over n} &=\int_{1^-}^x{\mathrm dA(t)\over t} \\ &=\int_1^x{\mathrm d(kt)\over t}+\int_1^x{\mathrm d[o(t)]\over t} \\ &\sim k\log x \end{aligned} $$

TravorLZH
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