# Is there any partial sums of harmonic series that is integer?

is there any partial sums of harmonic series that add up to an integer? partial sums not as trivial as the first term only i.e. 1, or the powers of 2 i.e the infinite geometric series for 2.

This might be trivially obvious for a subset of permutations of reciprocal of some integers.

PS: there is already another solved problem that states $\sum_{k=1}^n \frac{1}{k}$ is not an integer for all $n$, but my question allows removing finite or infinite subsets of terms, excluding 1 and geometric series for 2.

• It's not entirely clear to me what restrictions you have in mind on the subsets, but of course $1$ and $\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$ are integers... – Travis Willse Feb 7 '16 at 14:24
• yep, the second one, was that obvious to you? – Arjang Feb 7 '16 at 14:25
• I don't know about "obvious", but I know that example from another context. Note that we can use this to produce another one, $1 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}\left(\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}\right) = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{12} + \tfrac{1}{18} + \tfrac{1}{36}$, and we can iterate this process to produce infinitely many examples. – Travis Willse Feb 7 '16 at 14:35
• Or $\frac 12 +\frac 13+\frac 17 +\frac 1{42}$. Using Egyptian Fractions it isn't hard to generate examples. – lulu Feb 7 '16 at 14:36
• I have posted my question about Egyptian fractions...I expect the answer might have some interest. math.stackexchange.com/questions/1644638/… – lulu Feb 7 '16 at 16:19

One way to produce partial sums that add to $1$: Pick a natural number $N$ and write the Egyptian Fraction expansion for $\frac {N-1}N$. Then, unless $\frac 1N$ appears in that expansion, adding $\frac 1N$ to the list gets you an example.

For instance, starting from $N=347$ we get the example $$\frac 12 +\frac 13+\frac 17+\frac 1{48}+\frac 1{347}+\frac 1{10600}+\frac 1{154484400}=1$$ (happily, wolfram alpha knows how to compute Egyptian fractions).

This gives rise to a question which might be of some interest: For which natural numbers $N>1$ does $\frac 1N$ appear in the Egyptian fraction expansion of $\frac {N-1}N$? For instance $\frac 34=\frac 12+\frac 14$, or $\frac {11}{12}=\frac 12+\frac 13+\frac 1{12}$, or $\frac {83}{84}=\frac 12+\frac 13+\frac 17+\frac 1{84}$. Further searching also turned up the examples $N=3612, 6526884$. OEIS recognizes these as legs in integer Pythagorean triangles with some nice properties. Should add that my search was not at all comprehensive. Once I got the pattern from OEIS I checked higher terms in that sequence and confirmed that they had the desired property.

Edit: just to be clear, we're referring to that expansion obtained via the greedy algorithm. Egyptian expansions aren't unique and it's easy to produce examples using "non-greedy" expansions.

• Can't we just use the fact that "Every positive rational number can be represented by an Egyptian fraction." (Wikipedia) ? – Ant Feb 7 '16 at 16:11
• Not sure I follow. I mean the expansion obtained via the greedy algorithm...for present purposes, if $\frac 1N$ appears in that expansion, then it doesn't provide a viable answer to the question. – lulu Feb 7 '16 at 16:13
• Yes, but that's because you're looking at different ways to write $1$ as sum of fraction by first writing $(N-1)/N$ and then adding $1/N$ if it's not present already. But can't you just say "there are egyptian fraction that sum up to 1, to 2, 3 and so on for every integer, so we can find an infinite number of subsets of $\{1/n\}$ which sum up to an integer? – Ant Feb 7 '16 at 16:22
• Not familiar with Egyptian fractions for numbers like $2$ or greater integers. Pretty sure the Egyptians were happy leaving them as $2$ and such. But, yes. I was only talking about ways to get the partial sum $1$. – lulu Feb 7 '16 at 16:25
• Okay just wanted to check! I didn't know anything about egyptian fraction, only a quick look on wikipedia and I wanted to make sure I was not misunderstanding something :-) thanks – Ant Feb 7 '16 at 16:26

Since $\tfrac{1}{2} + \tfrac{1}{3} < 1$, any example must involve at least $3$ fractions, and it's not hard to see that $$\phantom{(\ast)} \qquad 1 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}, \qquad (\ast)$$ is the only example of length $3$.

One can also exploit identities among fractions of the form $\frac{1}{n}$ (called Egyptian fractions) like $$\tfrac{1}{m} = \tfrac{1}{m + 1} + \tfrac{1}{m (m + 1)} .$$ Indeed, applying this to the (inadmissible) decomposition $1 = \tfrac{1}{2} + \tfrac{1}{2}$ gives $(\ast)$, and applying this to $(\ast)$ itself gives both $$1 = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \tfrac{1}{12}$$ and lulu's example, $$1 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{7} + \tfrac{1}{42} .$$ There are precisely four more examples of length $4$ (for a total of $6$ such examples) and $72$ examples of length $5$. (See this calculator, which will generate all examples of a given length.)

You can also find infinite subseries of the harmonic series that converge to any positive real number $x$, whether $x$ is integer, rational, or irrational. The same is true for any divergent series $\sum a_n = \infty$ where $\lim_n a_n = 0$.

To find such a subseries $\sum_{i} a_{n_i} = x$, for each $j$, you choose $$n_j = \min\left\{n \in \Bbb N \mid n > n_{j-1} \text{ and } a_n < x - \sum_{j= 0}^{i - 1} a_{n_i}\right\}$$

For example, $$\frac 1{1} + \frac 1{2}+ \frac 1{3}+ \frac 1{7}+ \frac 1{43}+ \frac 1{1807} + ... = 2$$