Let $\left(a_n\right)$ be the following sequence: $$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many proofs for this one, but I think this is an elegant one. I'd like you to check it out, first because I consider it to be a nice and concise and elementary, and secondly because I'd like to make sure that there are no mistakes.

Let's get started: First of all, $a_n\geq 0 \ \forall \,n \in \mathbb{N}$. We write $a_n$ as

$$\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+2^k\left(\frac{\log\left(2\right)}{2}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}.$$

It is obvious that $\left(\frac{\log\left(n\right)}{n}\right)^k \to 0$. Let $\varepsilon >0$. Then $\exists\, n_0 \in \mathbb{N}$ such that $\forall \, n \geq n_0, \left(\frac{\log\left(n\right)}{n}\right)^k < \varepsilon$.

We have

\begin{align*} a_n&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k+ n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\ &=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+2^k+\dotsb +n^k} + \frac{n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\ &\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k}\\ & \qquad \qquad+ \frac{\varepsilon\left(1^k+\dotsb +\left(n_0-1\right)^k\right)+n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+\dotsb +n^k}\\ &\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k} + \frac{\varepsilon\left(1^k+2^k+\dotsb +n^k\right)}{1^k+2^k+\dotsb +n^k}\\ &=\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}+\varepsilon. \end{align*}

Now, by taking the limsup and the liminf as $n \to\infty$, and since

$$\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}\to 0,$$ we have

$$0 \leq \limsup\left(a_n\right) \leq \varepsilon \quad\text{and}\quad 0\leq \liminf\left(a_n\right) \leq \varepsilon.$$ But $\varepsilon$ was arbitrarily small, so


This is more of a discussion and not so much of a question :)

  • $\begingroup$ @DMcMor thanks for the edit! $\endgroup$ – Simply Beautiful Art Apr 16 '17 at 0:47
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    $\begingroup$ Perhaps you will also be interested in the Nørlund means, which generalizes this idea. $\endgroup$ – Sangchul Lee Apr 16 '17 at 1:25
  • $\begingroup$ @SangchulLee I'll check it out :) $\endgroup$ – JustDroppedIn Apr 16 '17 at 1:33

I have perhaps a more elegant proof:


The first step follows from the fact that $\frac ab<\frac cd$ if $c>a$ and $d<b$ for positive numbers $a,b,c,d$.

The limit then follows by letting $n^{k-1}=u^k$, which gives

$$\frac{\log^k(n)}{n^{k-1}}=\left[\frac k{k-1}\frac{\log(u)}u\right]^k$$

and the limit is then taken as given.

  • $\begingroup$ @SimplyBeautifulArt Your proof doesn't work if $k=1.$ $\endgroup$ – zhw. Apr 16 '17 at 2:26
  • $\begingroup$ @zhw.: Good catch! BTW for $k = 1$ he can sum the denominator to get $n(n + 1)/2$ and the numerator is bounded by $n\log n$ so that the limit is $0$ again. $\endgroup$ – Paramanand Singh Apr 16 '17 at 2:40
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    $\begingroup$ Also I don't get the fuss about the trivial thing $\log x < \sqrt{x}$ for large $x$. The inequality $\log x \leq x - 1$ is famous and replacing $x$ by $\sqrt{x}$ we get $\log x \leq 2 (\sqrt{x} - 1)$ and thus $(\log x)/x \to 0$ as $x \to \infty$. $\endgroup$ – Paramanand Singh Apr 16 '17 at 2:42
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen Apr 16 '17 at 5:09
  • $\begingroup$ @ParamanandSingh I and the OP obviously agree, though I just suppose some users take things differently than us. $\endgroup$ – Simply Beautiful Art Apr 16 '17 at 10:17

Fix $k\in \mathbb N.$ Let $a_n$ be the numerator, $b_n$ the denominator. Then $b_n \to \infty.$ Time to think about Stolz-Cesaro, which suggests we consider

$$\tag 1 \frac{a_{n+1}- a_n}{b_{n+1}- b_n} = \frac{\ln^k (n+1)}{(n+1)^k} = \left (\frac{\ln (n+1)}{n+1} \right )^k.$$

Since $[\ln (n+1)]/(n+1) \to 0,$ the right side of $(1)$ $\to 0.$ By Stolz-Cesaro, the limit of interest is $0.$

  • $\begingroup$ One of the many proofs I mentioned. Solid work :) $\endgroup$ – JustDroppedIn Apr 16 '17 at 2:26
  • $\begingroup$ this appears to be the slickest approach.+1 While I dislike the use of L'Hospital for functions, I like its cousin Cesaro-Stolz for sequences. $\endgroup$ – Paramanand Singh Apr 16 '17 at 2:43

Noting \begin{eqnarray} 0&<&\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}=\frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log i}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}\\ &\le& \frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}=\frac{(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}(\frac{i}{n})^k}, \end{eqnarray} and $$ \lim_{n\to\infty}\frac{\log n}{n}=0, \lim_{n\to\infty}\frac1n\sum_{i=1}^n(\frac{i}{n})^k=\int_0^1x^kdx=\frac1{k+1},$$ hence one has $$ \lim_{n\to\infty}\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}=0. $$

  • $\begingroup$ Another very nice solution :) $\endgroup$ – JustDroppedIn Apr 16 '17 at 21:37

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