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Prove that $\frac{1}{n} \sum_{k=2}^n \frac{1}{\log k}$ converges to $0.$

Okay, seriously, it's like this question is mocking me. I know it converges to $0$. I can feel it in my blood. I even proved it was Cauchy, but then realized that didn't tell me what the limit was. I've been working on this for an hour, so can one of you math geniuses help me?


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Your sum should start at $k=2$ ;) – N. S. Nov 3 '11 at 4:33
Yeah, it should. I changed it :D – badatmath Nov 3 '11 at 4:35
up vote 8 down vote accepted

In general, if $a_n\to 0$, then $\frac1n \sum_{k=0}^n a_k \to 0$ too.

(For any $\varepsilon>0$, find an $N$ such that $|a_n|<\varepsilon/2$ for all $n>N$, and then take enough terms beyond $N$ that they dominate whatever the terms before $N$ contribute to the average).

Even more generally (and as an easy consequence), whenever $a_n$ converges, $\frac1n \sum_{k=0}^n a_k$ converges to the same limit.

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Stolz Cezaro:

$$\lim \frac{1}{n} \sum_{k=2}^n \frac{1}{log k} = \lim \frac{1}{ \log (n+1)}$$

Edit Here is a direct proof:

$$0 \leq \frac{\sum_{k=2}^n \frac{1}{log k}}{n} = \frac{\sum_{k=2}^{\sqrt{n}} \frac{1}{log k}}{n} + \lim \frac{\sum_{k=\sqrt{n}}^n \frac{1}{log k}}{n} \leq \frac{\sqrt{n} \frac{1}{log 2}}{n} + \frac{(n-\sqrt{n})\frac{1}{\log \sqrt{n}}}{n} $$

Now each of the last two sequences converge to $0$, so squeeze it.

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Logarithm to any base is equal to a constant times logarithm to the base $2$. So let's work with base $2$. For convenience, write $\log n$ for $\log_2 n$.

We use a mild variant of the usual proof that $\sum \frac{1}{n}$ diverges.

Note that $\log 2$ and $\log 3$ are both $\ge 1$; $\log 4$, $\log 5$, $\log 6$, and $\log 7$ are all $\ge 2$; $\log 8$, $\log 9$, and so on up to $\log 15$ are all $\ge 3$, and so on. So we have: $$\frac{1}{\log 2}+\frac{1}{\log 3} \le \frac{1}{1}+\frac{1}{1}=2.$$ $$\frac{1}{\log 4}+\frac{1}{\log 5}+\frac{1}{\log 6}+\frac{1}{\log 7}\le \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2.$$

In the same way we can see that for any $m \ge 1$, $$\frac{1}{\log (2^{m})}+\frac{1}{\log (2^{m}+1)}+\cdots+\frac{1}{\log (2^{m+1}-1)}\le 2.$$ We conclude that $$\sum_{k=2}^{2^{m+1}-1} \frac{1}{\log k} <2m,$$ since up to $2^{m+1}-1$ we have $m$ chunks each of which has sum $<2$.

Now suppose that $2^{m} \le n \le 2^{m+1}-1$. Then $$\sum_{k=2}^n \frac{1}{\log k}\le \sum_{k=2}^{2^{m+1}-1} \frac{1}{\log k} <2m.$$ Thus $$\frac{1}{n}\sum_{k=2}^n \frac{1}{\log k} <\frac{2m}{2^{m}}.$$ Finally, it is well-known that $\displaystyle\lim_{m\to\infty} \frac{m}{2^m}=0$.

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