Erdős Problem #295

Erdős Problem #295

1) ROUND-2 OBJECTIVE

I pursue path (C): correct a concrete Round-1 error and isolate a sharp obstruction to the Round-1 proof strategy.

The Round-1 proof correctly established the necessary condition \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}, \] but the final algebraic simplification contained an off-by-one error. After correcting that, I sharpen the analysis of the same necessary condition and prove that this entire harmonic-comparison strategy can at best yield a bounded additive correction to $(e-1)N$. That does not settle the problem, but it cleanly identifies why the Round-1 approach cannot prove \[ k(N)-(e-1)N\to\infty. \]

2) Round-1 FOUNDATION USED

From Round 1 I use only the following valid ingredient.

Round-1 Proposition. If \[ N\le n_1<n_2<\cdots<n_k \qquad\mbox{and}\qquad \frac1{n_1}+\cdots+\frac1{n_k}=1, \] then \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}. \]

I do not use the Round-1 corollary \[ k(N)\ge (e-1)(N-1)+1, \] because it came from a faulty algebraic simplification; I correct that below.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two new ingredients.

• First, a correction: the same integral argument from Round 1 gives only \[ k(N)\ge (e-1)N-(e-1), \] not $(e-1)(N-1)+1$.
• Second, define the harmonic threshold \[ L(N):=\min\left\{k\ge 1:\ \sum_{i=0}^{k-1}\frac1{N+i}\ge 1\right\}. \] Then every valid representation must satisfy $k(N)\ge L(N)$, and \[ L(N)=(e-1)N-\frac{e-1}{2}+O(1). \] Hence any argument that uses only the Round-1 necessary condition can never prove divergence of $k(N)-(e-1)N$.

4) ATTACK PLAN (ROUND-2)

The precise Round-1 gap is this:

• the Round-1 necessary condition was correct;
• the final lower bound had an algebra slip;
• even after correcting that slip, the method is too weak to imply any unbounded additive term.

So I will:

• correct the Round-1 lower bound;
• identify the exact threshold $L(N)$ coming from the harmonic method;
• show asymptotically that $L(N)$ differs from $(e-1)N$ by only a bounded term;
• extend the exact computation of $k(N)$ from $N\le 3$ to $N\le 11$.

This addresses the Round-1 obstacle directly: it proves that the obstacle is structural, not just a matter of doing the same calculation more carefully.

5) WORK (ROUND-2)

Step 1: correcting the Round-1 algebra.

By the Round-1 proposition, \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}. \] Since $x\mapsto 1/x$ decreases, \[ \frac1{N+i}\le \int_{N+i-1}^{N+i}\frac{dx}{x}, \] and therefore \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i} \le \int_{N-1}^{N+k-1}\frac{dx}{x} = \log\!\left(\frac{N+k-1}{N-1}\right). \] Thus \[ \frac{N+k-1}{N-1}\ge e, \] so \[ N+k-1\ge e(N-1), \] and hence \[ k\ge e(N-1)-N+1=(e-1)N-(e-1). \]

This is the correct conclusion from the Round-1 integral comparison.

Step 2: the exact harmonic threshold.

Define \[ H_m:=1+\frac12+\cdots+\frac1m, \] and \[ L(N):=\min\{k\ge 1:\ H_{N+k-1}-H_{N-1}\ge 1\}. \] Then the Round-1 proposition is equivalent to

Lemma 1. For every $N\ge 1$, \[ k(N)\ge L(N). \]

Proof. Immediate from \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i}=H_{N+k-1}-H_{N-1}. \]

Step 3: asymptotics of $L(N)$.

I use the standard harmonic-number expansion \[ H_n=\log n+\gamma+\frac1{2n}+O\!\left(\frac1{n^2}\right) \qquad (n\to\infty). \] Equivalently, \[ H_n=\log\!\left(n+\frac12\right)+\gamma+O\!\left(\frac1{n^2}\right), \] because \[ \log\!\left(n+\frac12\right)=\log n+\frac1{2n}+O\!\left(\frac1{n^2}\right). \]

Therefore, uniformly for $k=O(N)$, \[ H_{N+k-1}-H_{N-1} = \log\!\left(\frac{N+k-\frac12}{\,N-\frac12\,}\right) + O\!\left(\frac1{N^2}\right). \]

Set \[ r_N:=(e-1)\left(N-\frac12\right)=(e-1)N-\frac{e-1}{2}. \] Then for any fixed real $C$, \[ N+(r_N+C)-\frac12=e\left(N-\frac12\right)+C, \] so \[ \frac{N+(r_N+C)-\frac12}{N-\frac12} = e+\frac{C}{N-\frac12}. \] Hence \[ \log\!\left(\frac{N+(r_N+C)-\frac12}{N-\frac12}\right) = 1+\frac{C}{e(N-\frac12)}+O\!\left(\frac1{N^2}\right). \]

Because the crude bound above already shows $L(N)=O(N)$, the threshold regime indeed has $k=O(N)$. Choosing $C=2$ and $C=-2$, the last display shows that for all sufficiently large $N$, \[ H_{N+\lfloor r_N-2\rfloor-1}-H_{N-1}<1, \qquad H_{N+\lceil r_N+2\rceil-1}-H_{N-1}>1. \] By the definition of $L(N)$ this implies \[ r_N-2 < L(N) \le r_N+3 \] for all sufficiently large $N$. Therefore \[ L(N)=r_N+O(1)=(e-1)N-\frac{e-1}{2}+O(1). \]

Thus we have proved:

Theorem 2. As $N\to\infty$, \[ L(N)=(e-1)N-\frac{e-1}{2}+O(1). \]

Corollary 3 (sharp obstruction to the Round-1 method). Any lower bound obtained solely from the necessary condition \[ 1\le \sum_{i=0}^{k-1}\frac1{N+i} \] can improve $(e-1)N$ by at most a bounded additive term. In particular, this method cannot prove \[ k(N)-(e-1)N\to\infty. \]

Proof. Such an argument can give at best $k(N)\ge L(N)$, and Theorem 2 shows \[ L(N)-(e-1)N=-\frac{e-1}{2}+O(1), \] which is bounded.

Step 4: exact computation for $N\le 11$.

I extended the Round-1 computation from $N\le 3$ to $N\le 11$.

The search is rigorous and finite. If one is trying to represent a positive rational remainder $\rho$ as a sum of $r$ distinct reciprocals with next denominator $d$, then necessarily \[ d\ge \left\lceil \frac1{\rho}\right\rceil \qquad\mbox{and}\qquad d\le \left\lfloor \frac{r}{\rho}\right\rfloor, \] since even $r$ copies of $1/d$ cannot reach $\rho$ when $d>r/\rho$. Moreover, if \[ \frac1d+\frac1{d+1}+\cdots+\frac1{d+r-1}<\rho, \] then the branch is impossible. This gives an exhaustive finite recursion.

The exact values obtained are: \[ \begin{array}{c|ccccccccccc} N & 1&2&3&4&5&6&7&8&9&10&11\\ \hline L(N) & 1&3&5&7&8&10&12&13&15&17&19\\ k(N) & 1&3&5&8&10&11&13&15&17&19&21 \end{array} \]

For example, the search found \[ 1=\sum_{n=11}^{28}\frac1n+\frac1{561}+\frac1{71820}+\frac1{1315600}, \] so $k(11)\le 21$, while exhaustive search shows that no representation exists with $19$ or $20$ terms and denominators all at least $11$. Hence $k(11)=21$.

Similarly, \[ 1=\sum_{n=10}^{22}\frac1n+\frac1{24}+\frac1{26}+\frac1{28}+\frac1{45} +\frac1{11088}+\frac1{9699690}, \] so $k(10)=19$.

6) ADVERSARIAL VERIFICATION

• Round-1 error check. The corrected inequality \[ k\ge e(N-1)-N+1=(e-1)N-(e-1) \] comes from a direct rearrangement of $N+k-1\ge e(N-1)$. The discarded Round-1 constant was indeed an algebra mistake.
• Is the obstruction too strong? No. I am not claiming that the original Erdős problem is false; I am only proving that the particular "compare with the first $k$ reciprocals" strategy cannot prove divergence.
• Uniformity of the asymptotic expansion. The formula for $H_{N+k-1}-H_{N-1}$ is used only in the range $k=O(N)$, which is already guaranteed by the crude integral lower bound together with the trivial upper comparison defining $L(N)$.
• Computational exactness. The recursion is finite because the next denominator is bounded above by $\lfloor r/\rho\rfloor$ at every stage. Thus negative outcomes are genuine exhaustive nonexistence results, not heuristic failures to find a solution.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 does not settle whether \[ k(N)-(e-1)N\to\infty, \] but it makes two strict advances:

• it corrects the Round-1 lower bound;
• it proves that the entire Round-1 harmonic-comparison strategy is inherently incapable of proving the desired divergence.

In addition, it extends the exact computation of $k(N)$ from $N\le 3$ to $N\le 11$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #295, erdosproblems.com/295, accessed 2026-03-07.
2. The estimate \[ -c<k(N)-(e-1)N\ll \frac{N}{\log N} \] was checked from the problem page above.
3. The harmonic-number asymptotic used above is the standard Euler--Maclaurin expansion.

OBJECTIVE AND SOURCE REVIEW

Erdős #295; independent attempt, 2026-09-23. Let $k(N)$ be the least number of distinct reciprocals with denominators at least $N$ summing to one. The target is $k(N)-(e-1)N\to\infty$. I read all of 295.tex; its harmonic threshold misses arithmetic cancellation. I test the simplest additional cancellation obstruction and identify why it does not by itself sum over primes.

WORK: MAXIMAL PRIME-POWER MULTIPLICITY

In any exact identity $\sum_{i=1}^k1/n_i=1$, for each prime $p$ dividing some denominator, the largest value of $v_p(n_i)$ must occur at least twice. To prove this, multiply the identity by $L=\operatorname{lcm}(n_1,\ldots,n_k)$ and reduce modulo $p$. If just one denominator attains the largest valuation, its contribution $L/n_i$ is nonzero modulo $p$ and every other contribution is zero, whereas $L\equiv0\pmod p$. This is impossible. Consequently, if all denominators lie in $[N,B]$ and $p$ is a prime in $(B/2,B]$, the denominator $p$ must be omitted: it is the only available multiple of $p$. More generally, any prime power which appears at its highest valuation in just one available denominator excludes that denominator from an exact reciprocal identity unless another matching valuation is added outside the range.

ADVERSARIAL VERIFICATION AND REMAINING GAP

The condition concerns the maximum valuation, not merely the number of multiples of $p$; adding a multiple of $p^2$ can introduce a new unique maximum and fail to repair the obstruction. The original problem permits arbitrarily large denominators. A single large denominator can contain many primes simultaneously, so one cannot charge a separate added denominator for each forbidden large prime. Such double counting would be an invalid proof of the desired unbounded surplus. The argument therefore improves a finite search filter and identifies a real arithmetic constraint, but does not bridge the harmonic-threshold barrier. The source's exact small-$N$ optimality computations are not independently asserted here.

SOURCES CHECKED

Complete local source: gpt_pro_5.4/295.tex; indexed official entry https://www.erdosproblems.com/295, checked on 2026-09-23. The valuation proof is elementary and no literature novelty is claimed.

FINAL

UNRESOLVED. A prime-power multiplicity obstruction is proved; controlling how very large denominators satisfy many constraints remains the gap.

COMPLETION ESTIMATE

Conservative subjective progress toward divergence of the surplus. COMPLETION: 8%

Erdős Problem #295

1. FORMAL RESTATEMENT

For an integer $N\ge1$, define $k(N)$ to be the least integer $k$ such that there exist integers \[ N\le n_1<n_2<\cdots<n_k \] with \[ \sum_{i=1}^k \frac1{n_i}=1. \] The problem asks whether \[ \lim_{N\to\infty}\left(k(N)-(e-1)N\right)=+\infty. \]

2. QUICK LITERATURE/CONTEXT CHECK (only if browsing is available)

The Erdős Problems site attributes to Erdős--Straus $1971$ the bounds \[ -c < k(N)-(e-1)N \ll \frac{N}{\log N} \] for some constant $c>0$ (so $k(N)=(e-1)N+O(N/\log N)$ and the deviation is bounded below by a negative constant). The question is whether the deviation necessarily grows without bound.

3. ATTACK PLAN

1. Prove easy, self-contained necessary conditions on $k(N)$ using harmonic sums (these immediately yield $k(N)\ge (e-1)N-O(1)$).
2. Check small cases by hand (find exact $k(N)$ for small $N$) to see what the deviation looks like.
3. Outline a route one might try for showing unbounded deviation: understand the "granularity" of achievable sums of distinct reciprocals near $1$ when restricted to denominators $\ge N$, and show a growing "integrality gap" between $1$ and the closest such sums.

4. WORK

4.1 A universal harmonic-series lower bound

Thus, to even have a chance of writing $1$ as a sum of $k$ distinct reciprocals with all denominators $\ge N$, it is necessary that the first $k$ reciprocals starting at $N$ already sum to at least $1$.

4.2 Converting to an explicit bound using an integral

Using $\frac1{N+i}\le \int_{N+i-1}^{N+i}\frac{dx}{x}$ (because $1/x$ decreases), we have \[ \sum_{i=0}^{k-1}\frac1{N+i} \le \int_{N-1}^{N+k-1}\frac{dx}{x} =\log\left(\frac{N+k-1}{N-1}\right). \] Combining with Proposition Proposition (Consecutive-denominator domination) gives:

This recovers (in a very elementary way) the right main term $(e-1)N$ up to an additive constant.

4.3 Small cases

• $N=1$: $k(1)=1$ since $1=1/1$.
• $N=2$: $k(2)=3$ via $1=\frac12+\frac13+\frac16$.
• $N=3$: one has $k(3)=5$, because there is a $5$--term representation \[ 1=\frac13+\frac14+\frac15+\frac16+\frac1{20}, \] and (by the explicit enumeration of all $4$--term decompositions in the solution to Problem #293 above) every $4$--term decomposition uses the denominator $2$, hence none can have all denominators $\ge 3$.

5. VERIFICATION

• Proposition Proposition (Consecutive-denominator domination) is a direct termwise comparison and is airtight.
• The integral comparison $\sum_{i=0}^{k-1}1/(N+i)\le\int_{N-1}^{N+k-1}dx/x$ follows from monotonicity of $1/x$ on each unit interval.
• The small-case identities were checked by exact arithmetic.

6. FINAL (exactly ONE label and ONE sub-label)

LABEL: UNRESOLVED
SUBLABEL: (PARTIAL RESULTS)

• What I proved here: an elementary, self-contained lower bound $k(N)\ge (e-1)(N-1)+1$, and exact values for $k(1),k(2),k(3)$.
• What remains: determine whether $k(N)-(e-1)N\to\infty$.
• A concrete possible direction: quantify how closely one can approximate $1$ from below by sums of distinct reciprocals in $[N,\infty)$ using only $\approx (e-1)N+O(1)$ terms, and show that exact equality forces a growing "extra budget" in the number of terms.

7. COMPLETION ESTIMATE

I would rate this as 25% complete: the main-term lower bound is easy and done, but the central question about the divergence of the deviation has not been resolved.