# A convergent, but large, sum of integer reciprocals

The sums of the reciprocals of monotonic integer sequences may be convergent or diverge to infinity. (For example) Wikipedia gives examples of both. So the sum of the reciprocals of the prime numbers diverges, whereas the sum of the reciprocals of the factorials converges to e.

I note that all of the examples given of convergent sums, converge to a fairly small value. The largest given by Wikipedia is the sum of the Kempner series (reciprocals of all integers that don't include '9' in their decimal representation). It's sum is approximately 22.9 This is still a fairly small value.

While it would be possible to produce sequences that have larger sums by trival modifications of existing sequences (all numbers that don't include '99') or by combining sequences (reciprocals of numbers that are either factorials or square numbers or...) I would like to find a monotonic infinite sequence of integers that: isn't a simple manipulation or combination of another sequence or sequences, and converges to a notably large number. (And by "large" I mean at least 22.9, but the larger the better)

My thoughts on finding such a sequence: If there were a natural sequence of integers that included all numbers, to a certain point, and then became sparse, it might satisfy my conditions. I thought of "the reciprocals of all integers $$n$$ such that $$\pi(n) < \mathrm{li}(n)$$ However I suspect this sequence is divergent.

It would, be possible to combine a finite sequence with a convergent infinite one (all numbers that are less than a googol or in form $$2^n$$) but this would fail my criteria of not being a combination of other sequences.

• The sum of the reciprocals of the integers $n$ with $\pi(n)<li(n)$ could well be convergent. Not sure whether there is hope to decide this. Commented Aug 8 at 9:11
• The sum of the reciprocals of all integers $n$ in a given set is asymptotically $\log x$ times the logarithmic density of that set (by definition, assuming the logarithmic density is positive). Assuming the Riemann hypothesis and another condition, Rubinstein and Sarnak proved that the logarithmic density of $\{n\in\Bbb N\colon \pi(n) < \mathop{\rm li}(n)\}$ is indeed positive, so the corresponding sum of reciprocals diverges. Commented Aug 8 at 9:13
• The sum of 1/n grows only as ln n. So a convergent sum of 1,000,000 would require at least about exp(1,000,000) terms which is quite a lot. Commented Aug 8 at 18:08
• Let $d_b(n)$ be the number of digits in $n$ written in integer base $b$ so $d_b(n)=\lfloor \log_b(n)+1\rfloor$. Consider $\sum\limits_{n=1}^\infty \frac1{n\, d_b(n)^2}$, which converges for any $b$, but to an arbitrarily large value as $b$ increases. If you think the square is arbitrary, try $\sum\limits_{p\text{ prime}} \frac1{p\, d_b(p)}$ instead. Commented Aug 9 at 0:20

For any positive real number $$\alpha$$, the greedy Egyptian fraction algorithm recursively defines an increasing sequence $$n_1,n_2,\dots$$ as follows: take $$n_j$$ to be the smallest integer exceeding $$n_{j-1}$$ for which $$\sum_{i=1}^j \frac1{n_i} < \alpha.$$ (By convention, $$n_0=0$$.) [The standard greedy algorithm uses $$\le\alpha$$ instead of $$<\alpha$$ and results in a finite sequence if $$\alpha$$ is rational; but the $$<$$ version is also well known and always produces an infinite sequence $$(n_j)$$ (for example, when $$\alpha=1$$ the resulting sequence is Sylvester's sequence).] In this way we produce a convergent sequence of reciprocals of positive integers with $$\sum_{i=1}^\infty\frac1{n_i} = \alpha,$$ for any positive real number $$\alpha$$ we choose.

• That works, but is somehow unsatisfying. Commented Aug 8 at 11:02
• That's a reasonable reaction :) The values of series of this type are clearly less special than they might first appear Commented Aug 8 at 16:43
• Could you provide a link to a definition of the < version of this algorithm? Or does it suffice to choose an irrational k such that k < $\alpha$ ? Commented Aug 8 at 19:04
• I tried to make the definition here self-contained and agnostic to the rationality of $\alpha$; is there a way the algorithm can be clarified? Commented Aug 8 at 21:35
• @GregMartin ahh, my brain woke up. Just start with $\frac{n_1} < \alpha}$ and go on from there. Commented Aug 9 at 15:10

How about $$\{\frac{1}{\lfloor n^r\rfloor}:n\in\mathbb N\},$$ where $$r$$ is a real number? As long as $$r>1$$ it will always be finite but as $$r\to 1$$ the sum will approach infinity, so you can make it as large as you want.

• Yes, but I think I can express what is "unsatisfying" about these examples (and what, conversely, is satisfying about the Kempner series). Both this and Greg's answer are recipes for producing a series with an arbitrarily large sum. But the actual specific series that gives my "large" sum is really just a "trivial modification" of one with a small sum. In this case you could consider the case with r=2 as a particular series and r=1+\epsilon as a modification. Similarly in Gregs example, you pick a target and then build a sequence to get to that target Commented Aug 9 at 17:48
• In contrast, the Kempner series is the simplest of it's kind (in some way) I can make series that have arbitrarily large sums by modifying Kempner And I see this example as being essentially the same, a modification of the 1/n^2 sequence. But I think the answer is useful anyway. Commented Aug 9 at 17:52