li. Ant on a Rubber Rope

Problem. A rubber rope is 1 km long. An ant starts at one end and crawls at 1 cm/s toward the other. After every second the rope is stretched uniformly by 1 km (the ant is carried along). Does the ant ever reach the end?

Answer. Yes. Measure progress as the fraction of the rope covered: stretching does not change it, and in second $k$ the ant adds $\frac{1\text{cm}}{k\text{km}}=\frac{1}{10^{5}k}$. After $n$ seconds the fraction is $\frac{H_n}{10^5}$, and the harmonic series diverges, so the ant arrives when $H_n\ge 10^5$, after about $e^{10^5-\gamma }\approx 10^{43429}$ seconds (the universe is about $4\cdot 10^{17}$ s old).

Model	Fraction covered	Arrival
discrete: length $a$, $+b$ after each second, speed $c$	${\sum }_{k=0}^{n-1}\frac{c}{a+kb}$	first $n$ with the sum $\ge 1$
$a=b$ (classic)	$\frac{c}{a}{H}_n$	first $n$ with $H_n\ge a/c$, $n\approx e^{a/c-\gamma }$
continuous: length $L+vt$, speed $u$	$\frac{u}{v}\ln (1+\frac{vt}{L})$	$T=\frac{L}{v}(e^{v/u}-1)$, finite for every $u>0$
first $n$ with $H_n\ge K$, $K=\mathrm{1..10}$	$1,4,11,31,83,227,616,1674,4550,12367$	ratio tends to $e$
rope stretched geometrically (length $L{q}^t$, $q>1$)	$\frac{u}{L}\sum q^{-k}$ converges	may never arrive: needs $u\ge L(1-1/q)$ in the discrete version

// Ant at speed c on a rope of length a that is stretched uniformly by b after each second.
ll antArrival(double a, double b, double c) {        // first n with sum_{k<n} c / (a + k b) >= 1
    double f = 0;
    for (ll n = 1;; n++) { f += c / (a + (n - 1) * b); if (f >= 1) return n; }
}
 
ll firstHarmonicAtLeast(long double K) {             // smallest n with H_n >= K, for K up to about 30
    const long double g = 0.57721566490153286060651209L;  // Euler-Mascheroni constant
    auto H = [&](ll n) -> long double {
        if (n < 64) { long double s = 0; for (ll i = 1; i <= n; i++) s += 1.0L / i; return s; }
        long double x = n; return logl(x) + g + 1 / (2 * x) - 1 / (12 * x * x) + 1 / (120 * x * x * x * x);
    };
    ll lo = 1, hi = 1;
    while (H(hi) < K) hi *= 2;
    while (lo < hi) { ll mid = lo + (hi - lo) / 2; if (H(mid) >= K) hi = mid; else lo = mid + 1; }
    return lo;
}

Proof: the fraction argument, divergence and the arrival times

Fraction. Uniform stretching multiplies every distance from the start, the ant’s position and the rope length alike, so the covered fraction $f$ is unchanged by stretching. During second $k$ (rope length $a+(k-1)b$) the ant adds $c/(a+(k-1)b)$ to $f$. Summing gives the table formula; for $a=b$ it is $\frac{c}{a}{\sum }_{k=1}^n\frac{1}{k}=\frac{c}{a}{H}_n$.

Divergence. For $k\ge 1$, $a+kb\le (a+b)k$, so $\frac{c}{a+kb}\ge \frac{c}{a+b}\cdot \frac{1}{k}$, and $\sum \frac{1}{k}$ diverges (group terms $2^j<k\le 2^{j+1}$: each group sums to more than $\frac{1}{2}$). So $f$ eventually exceeds 1 whatever $a,b,c>0$. With $H_n=\ln n+\gamma +\frac{1}{2n}-O(n^{-2})$, $H_n\ge K$ first happens at $n\approx e^{K-\gamma }$; for $K=10^5$, ${\log }_{10}n\approx (10^5-0.577)/\ln 10\approx 43429$.

Continuous. $f^{\prime }(t)=\frac{u}{L+vt}$, so $f(t)=\frac{u}{v}\ln \frac{L+vt}{L}$, which equals 1 at $t=\frac{L}{v}(e^{v/u}-1)$. Geometric stretching: the increments $\frac{u}{L{q}^k}$ sum to $\frac{u}{L}\cdot \frac{q}{q-1}$, so the ant arrives iff this reaches 1, i.e. $u\ge L(1-1/q)$.

The code. antArrival accumulates exactly the increments. firstHarmonicAtLeast binary searches with $H_n$ computed directly for small $n$ and by the Euler–Maclaurin expansion (error below $n^{-6}$) otherwise; $H$ is increasing, so the search is valid while the long double error stays below the gap $H_n-K$ at the answer, which holds for $K\lesssim 30$. Checked: antArrival against a step-by-step simulation of positions and lengths, firstHarmonicAtLeast against direct summation for $K\le 15$.

Related notes

• Harmonic Lemma and Floor Sums
• Jeep Problem (another harmonic-series puzzle)
• Probability and Expectation (coupon collector also gives $n{H}_n$)