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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.

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    $\begingroup$ 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... $\endgroup$ – Travis Willse Feb 7 '16 at 14:24
  • $\begingroup$ yep, the second one, was that obvious to you? $\endgroup$ – Arjang Feb 7 '16 at 14:25
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    $\begingroup$ 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. $\endgroup$ – Travis Willse Feb 7 '16 at 14:35
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    $\begingroup$ Or $\frac 12 +\frac 13+\frac 17 +\frac 1{42}$. Using Egyptian Fractions it isn't hard to generate examples. $\endgroup$ – lulu Feb 7 '16 at 14:36
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    $\begingroup$ I have posted my question about Egyptian fractions...I expect the answer might have some interest. math.stackexchange.com/questions/1644638/… $\endgroup$ – lulu Feb 7 '16 at 16:19
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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.

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    $\begingroup$ Can't we just use the fact that "Every positive rational number can be represented by an Egyptian fraction." (Wikipedia) ? $\endgroup$ – Ant Feb 7 '16 at 16:11
  • $\begingroup$ 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. $\endgroup$ – lulu Feb 7 '16 at 16:13
  • $\begingroup$ 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? $\endgroup$ – Ant Feb 7 '16 at 16:22
  • $\begingroup$ 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 $1$. $\endgroup$ – lulu Feb 7 '16 at 16:25
  • $\begingroup$ 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 $\endgroup$ – Ant Feb 7 '16 at 16:26
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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.)

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

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