For $\lambda\ge1/(n-1)$ just repeatedly jump the leftmost flea over the rightmost flea. Now assume that $\lambda<1/(n-1)$. Let the positions after the $k^{th}$ jump be $x_1<x_2<\cdots<x_n$. If $x_i<x_j$, then jumping is equivalent to \[(x_i,x_j)\mapsto (x_j,x_j+\lambda(x_j-x_i)).\]Letting $M_k,s_k$ denote the maximum and sum (respectively) of the positions after $k$ jumps, we see that \[M_{k+1}-M_k \le x_j+\lambda(x_j-x_i)-x_n \le \lambda(x_j-x_i)\]while \[s_{k+1}-s_k = x_j+x_j+\lambda(x_j-x_i)-x_i-x_j = (1+\lambda)(x_j-x_i),\]whence summing up yields \[\frac{M_l-M_0}{s_l-s_0} \le \frac{\lambda}{1+\lambda} = c < \frac{1}{n}\]for all $l\ge1$. Then \[M_l-M_0 \le c(s_l-s_0) \le c(nM_l-s_0)\implies M_l \le \frac{M_0-cs_0}{1-cn}.\]

Assume $ \lambda < \frac{1}{n-1} $. Then for a position of the fleas $ f=(f_1,...,f_n) $ with $ f_1 \ge ... \ge f_n $, define $ M(f) = f_1 + \left( \frac{\lambda}{1-(n-1)\lambda} \right) \cdot \left( \sum_{k=2}^n (f_1-f_k) \right) $. Clearly if $ f_n > M $ then $ M(f) \ge f_1 > M $. Thus, if the fleas were able to jump to infinity, then $ M \rightarrow \infty $. However, $ M $ is a monovariant! (It never increases). Indeed, if $ f_1 $ doesn't change then $ M $ decreases strictly, and if $ f_1 $ changes, $ M $ is easily computed to stay the same. Thus, we're done. If $ \lambda \ge \frac{1}{n-1} $, then by making the last one jump over the first one, we compute that the average distance between consecutive fleas never decreases, and $ f_n$ always increases, thus the fleas jump to infinity.

Hello! I was wondering if the following solution is correct? Here we replace lambda with c: After each step k, denote x(k)(1), x(k)(2), ..., x(k)(n) in increasing orders to be the positions of the fleas. Hence x(n)(k) is the maximum position. Note that the optimal way to increase the maximum position is by jumping the leftmost fly over the rightmost fly. Now, denote P(k) to be the maximum value of the positions after k steps (essentially the same as x(n)(k)). Note that P(k+1) - P(k) = c(x(n)(k) - x(n)(0)). After some algebra we get that c[(x(n)(k) - x(n)(1))] = [x(0)(n) - x(0)(1)][(n-1)(c)]^k, where x(0)(i) denotes the initial positions. Finally, let d = c(n-1). By collapsing the sum, we get that P(k) = (d^(k-1)+d^(k-2)+...+1)[x(0)(n) - x(0)(1)] = [(d^k - 1)/(d - 1)] * [x(0)(n) - x(0)(1)]. Note that [x(0)(n) - x(0)(1)] is constant, and we want that P(k) is unbounded as k approaches infinity. For all c < 1/(n-1), we see that P(k) is bounded above as k approaches infinity by 1/(1-d) * [x(0)(n) - x(0)(1)], and for c >= 1/(n-1) P(k) is unbounded, so we can move the fleas! Hence the desired answer is all c (lambda) >= 1/(n-1). Thanks!

Could anyone help?

\bump? Thanks

Write $n = N+1$; we claim the answer is \[ \boxed{\left\{\lambda\colon \lambda \geq \frac 1N\right\}}. \]Place the points on the number line; let $p$ denote the coordinate of point $P$. For a configuration $x_1\leq x_2\leq \cdots \leq x_{N+1}$ of points $X_1,X_2,\ldots, X_{N+1}$, define its cap by $x_{N+1}$ and its score by $X_1X_{N+1} + X_2X_{N+1} + \cdots + X_NX_{N+1}$. Notice now that we can get all the points past $M$ if and only if we can get the cap to exceed $m$; only if clearly holds, and for if just notice that if $X_{N+1}$ is past $M$, we just reflect everything over $X_{N+1}$ and win. So, we just want to find $\lambda$ so that the cap can get arbitrarily large. Lemma 1: $\lambda \geq \frac 1N$ all work. Proof: First, note that we can get a configuration in which no two fleas are at the same point. To do this, first take the leftmost point and reflect over the rightmost point. Then, we have a point strictly to the right of all other points. Suppose the points are $P_1,P_2,\ldots, P_{N+1}$, respectively. Then, at each point, reflect the leftmost point over the rightmost one $N$ times, until we have $N+1$ points all at distinct places. Suppose our current configuration is $X_1,X_2,\ldots, X_{N+1}$, and let $d_1 = X_1X_2$, $d_2 = X_2X_3, \ldots$, and write $d = \min(d_i)>0$. Reflect $X_1$ over $X_{N+1}$. Then, the configuration becomes $X_2,X_3,\ldots, X_{N+1}, X_1'$, and the pairwise distances become $d_2,d_3,\ldots, d_N, \lambda(d_1+d_2+\cdots+d_N)$, which are again all $\geq d$. So, at each step, we know that the cap has coordinate increasing by at least $d$ every time, so we can get the cap to be arbitrarily large. $\Box$ Lemma 2: $\lambda < \frac 1N$ all fail. Proof: Define the value $V$ of a configuration to be \[ V = C + \frac{\lambda S}{1-N\lambda}, \]where $C$ is the cap and $S$ is the score, respectively. We claim that the value is nonincreasing. Indeed, suppose we have a configuration $X_1,X_2,\ldots, X_{N+1}$, where $x_1\leq x_2\leq \cdots \leq x_{N+1}$. Suppose we reflect $X_i$ over $X_j$ with $i<j$. Write $\alpha = X_iX_j$, and $\beta =X_jX_{N+1}$. If the cap does not change, then clearly the score decreases, since $X_i$ moves closer to $X_{N+1}$ and the other distances do not change. Otherwise, we see that the cap increases by $\lambda \alpha - \beta$. Furthermore, since we went from $X_1,X_2,\ldots, X_{N+1}$ to $X_1,X_2,\ldots, X_{i-1},X_{i+1},\ldots, X_{N+1},Q$, where $Q$ is where $X_i$ jumped to, then the score changes from \[ \sum_{k=1}^N x_{N+1}-x_k = Nx_{N+1} - \sum_{k=1}^N x_k \]to \[ \sum_{\substack{k=1\\ k\neq i}}^{N+1}q - x_k = Nq - \sum_{\substack{k=1\\k\neq i}}^{N+1} x_k, \]so we compute that $\Delta S = N(q-x_{N+1}) - x_{N+1}+x_i = N(\lambda\alpha-\beta) - (\alpha+\beta)$. So, \[ \Delta V = \Delta C + \frac{\lambda \Delta S}{1-N\lambda} = (\lambda\alpha-\beta) + \frac{\lambda (N(\lambda\alpha-\beta)-\alpha-\beta)}{1-N\lambda} \leq \lambda\alpha + \frac{\lambda(N\lambda-1)\alpha}{1-N\lambda} = 0. \]So, $V$ is indeed nonincreasing. So, if $V_0$ is the initial value, since $S\geq 0$ always we have that \[ C\leq V \leq V_0 \]implying that $m = V_0$ indeed works, and that we cannot move all the fleas to the right of $M$. $\Box$ Thus, with both lemmas proven, we may conclude.

The answer is $\lambda \ge \frac{1}{n-1}$. We change the problem be replacing the fleas by bowling balls $B_1$, $B_2$, \dots, $B_n$ in that order. Bowling balls aren't exactly great at jumping, so the operation now works as follows. Choose any indices $i < j$. Then ball $B_i$ moves to $B_{i+1}$'s location, $B_{i+1}$ moves to $B_{i+2}$'s location, and so on; until $B_{j-1}$ moves to $B_j$'s location, Finally, $B_j$ moves some distance at most $\lambda \cdot |B_j B_i|$ forward, but it may not pass $B_{j+1}$. Claim: If $\lambda < \frac{1}{n-1}$ the bowling balls have bounded movement. Proof. Let $a_i \ge 0$ denote the initial distance between $B_i$ and $B_{i+1}$, and let $\Delta_i$ denote the distance travelled by ball $i$. Of course we have $\Delta_1 \le a_1 + \Delta_2$, $\Delta_2 \le a_2 + \Delta_3$, \dots, $\Delta_{n-1} \le a_{n-1} + \Delta_n$ by the relative ordering of the bowling balls. Finally, distance covered by $B_n$ is always $\lambda$ times distance travelled by other bowling balls, so \begin{align*} \Delta_n &\le \lambda \sum_{i=1}^{n-1} \Delta_i \le \lambda \sum_{i=1}^{n-1} \left( \left( a_i + a_{i+1} + \dots + a_{n-1} \right) + \Delta_n \right) \\ &= (n-1)\lambda \cdot \Delta_n + \sum_{i=1}^{n-1} i a_i \end{align*}and since $(n-1)\lambda > 1$, this gives an upper bound. $\blacksquare$ Claim: When $\lambda \ge \frac{1}{n-1}$, it suffices to always jump the leftmost flea over the rightmost flea. Proof. If we let $x_i$ denote the distance travelled by $B_1$ in the $i$th step, then $x_i = a_i$ for $1 \le i \le n-1$ and $x_i = \lambda(x_{i-1} + x_{i-2} + \dots + x_{i-(n-1)})$. In particular, if $\lambda \ge \frac{1}{n-1}$ then each $x_i$ is at least the average of the previous $n-1$ terms. So if the $a_i$ are not all zero, then $\{x_{n}, \dots, x_{2n-2}\}$ are all positive and thereafter $x_i \ge \min \left\{ x_n, \dots, x_{2n-2} \right\} > 0$ for every $i \ge 2n-1$. So the partial sums of $x_i$ are unbounded, as desired. $\blacksquare$

I claim the answer is $\lambda \ge \frac{1}{n-1}$. To see this works, consider the strategy of hopping the leftmost flea over the rightmost. After $n-1$ moves, all fleas will now be at distinct positions, and define $m$ to be the minimum distance between two consecutive fleas. Now, the minimum distance is always at least $m$, since the distance made by hopping is the average of all the other distances. So, every hop, the configuration's leftmost point moves forward at least $m$, and all fleas will eventually pass $M$. If $\lambda <\frac{1}{n-1}$, define a point $P$, which is left of all the fleas, and call the distance from the $n$th rightmost flea to $P$ $d_n$. Then, call the leakiness of the configuration to be $d_1-\lambda(d_2+\ldots+d_n)$. If $\lambda<\frac{1}{n-1}$, this number is clearly always positive. Now, note that if a flea jumps but does not go past the 1st flea, then the leakiness must decrease. If the $i$th flea jumps over the $j$th, and actually becomes the first flea, then the leakiness changes by $-(1+\lambda)d_1+(d_j-d_i)\lambda+d_j+\lambda d_i=(1+\lambda)(d_j-d_1)$. As $d_j\le d_1$, this value is nonpositive, so the leakiness doesn't increase. The minimum leakiness required to get across $M$ is $(1-(n-1)\lambda)PM$, so if we move $M$ sufficiently far to the right, this quantity becomes larger than the leakiness of the initial configuration, as desired.

I would like to point out that the problem solver should notice that the rightmost guy is special. We want the rightmost guy to the as far to the right as possible and all others as far left as possible. Then just from this we can make some monovariant.

The answer is $\lambda\ge\frac1{n-1}$. Proof of lower bound. We will use this obvious strategy --- the leftmost flea always jumps over the rightmost flea. Consider the nondecreasing sequence $(x_i)_{i\ge1}$ with $0<x_1\le\cdots\le x_n$ the starting coordinates of the fleas; for each $k>n$, let $x_k=x_{k-1}(\lambda+1)-x_{k-n}\lambda$. After $k$ moves, the fleas have coordinates $x_{k+1}$, $\ldots$, $x_{k+n}$. I will prove this sequence is unbounded. Summing over $x_k-x_{k-1}=\lambda(x_{k-1}-x_{k-n})$, we have \[x_k-x_n=\lambda\left(x_{k-1}+\cdots+x_{k-(n-1)}\right)-\lambda(x_1+\cdots+x_{n-1}).\]In other words, $x_k-\lambda\left(x_{k-1}+\cdots+x_{k-(n-1)}\right)$ is a fixed constant $C>0$. Then since $x_{k-(n-1)}\le x_{k-i}$ for $0\le i\le n-2$, we have \begin{align*} \frac{x_k+\cdots+x_{k-(n-2)}}{n-1}&\ge\frac{x_k+\cdots+x_{k-(n-1)}}n\\ &=\frac{C+(\lambda+1)\left(x_{k-1}+\cdots+x_{k-(n-1)}\right)}n\\ &\ge\frac Cn+\frac{x_{k-1}+\cdots+x_{k-(n-1)}}{n-1}, \end{align*}so $x_{k-1}+\cdots+x_{k-(n-1)}$ is unbounded, and thus so is $(x_i)_{i\ge1}$. Proof of upper bound. At some point in time, let the fleas have coordinates $0<x_1\le x_2\le\cdots\le x_n$, and let $I=x_n-\lambda(x_1+\cdots+x_{n-1})$. Claim: $I$ is nonincreasing. Proof. Let $x_i$ jump over $x_j$ to $x_i'$, so that $x_i'=x_j(\lambda+1)-x_i\lambda$, and let $I$ map to $I'$. If $x_i'<x_n$, then \[I'=I-\lambda(x_i'-x_i)=I-\lambda(\lambda+1)(x_j-x_i)<I.\]Otherwise $x_i'\ge x_n$, so $x_i'$ is the new maximum and \begin{align*} I'&=I+x_i'-x_n-\lambda(x_n-x_i)\\ &=I-x_n(\lambda+1)+(x_i'+\lambda x_i)\\ &=I-(x_n-x_j)(\lambda+1)\le I, \end{align*}which proves the claim. $\blacksquare$ However if at some point $x_n>M$, then \begin{align*} I&=x_n-\lambda(x_1+\cdots+x_{n-1})\\ &>x_n-\lambda(x_n+\cdots+x_n)\\ &=x_n\big[1-(n-1)\lambda\big]\\ &>M\big[1-(n-1)\lambda\big], \end{align*}which is larger than $I$ for sufficiently large $M$.

We want to find all $\lambda$ for which it is impossible to move all fleas infinitely rightward. I claim our answer is all $\lambda \geq \tfrac{1}{n-1}$. When $\lambda \geq \tfrac{1}{n-1}$, we can simply let the leftmost flea jump over the rightmost one on each move. After $n-1$ moves, we will have $n$ fleas in positions $p_1 < p_2 < \ldots < p_n$. If we let $\delta = \min{(p_{i+1} - p_i)}$ for $i \in [1, n-1]$, then it is easy to see that after each move, the distance between the leftmost and rightmost fleas is nondecreasing and thus so is $\min{(p_{i+1} - p_i)}$, and that thus the new leftmost flea's position is at least $\delta$ to the right of the old one. Hence the leftmost flea can go arbitrarily far to the right by moving $\delta$ each time, and thus so can the rest of the fleas. When $\lambda < \tfrac{1}{n-1}$, I claim that the position of the rightmost flea is bounded. Indeed, suppose that after some $\ell$ moves, the $n$ fleas are at positions\[q_1 \leq q_2 \leq \ldots \leq q_n\]with largest $m_{\ell} = q_n$ and sum $s_{\ell}$. If we were to perform a move on fleas on $q_i < q_j$, which must exist since all fleas can never be on the same point, then: the maximum changes from $m_k$ to $m_{k+1}$ by at most $\lambda(x_j - x_i)$, which happens when $j = n$. the sum changes from $s_k$ to $s_{k+1}$ by exactly $(x_j + [x_j + \lambda(x_j - x_i)]) - (x_i + x_j) = (1 + \lambda)(x_j - x_i)$ Therefore, after any number of moves $K$, we see that\[m_K \leq m_0 + (\lambda y_1 + \ldots + \lambda y_K) = m_0 + \lambda(y_1 + \ldots y_K)\]and\[s_K = s_0 + ([1 + \lambda]y_1 + \ldots + [1 + \lambda]y_K) = s_0 + (1 + \lambda)(y_1 + \ldots + y_K)\]where $y_i$ is the positive difference in the positions of the fleas moved in the $i$th move. Thus, we see that the ratio $\tfrac{\Delta m}{\Delta s}$ is at most $\tfrac{\lambda}{1 + \lambda}$ after any number of moves, and in fact we can solve\[\frac{m_K - m_0}{s_K - s_0} \leq \frac{\lambda}{1 + \lambda} \implies m_K \leq m_0 + \frac{\lambda}{1 + \lambda}\left(s_K - s_0\right) \leq m_0 + \frac{\lambda}{1 + \lambda}(nm_K - s_0)\]and obtain that $m_K \leq \tfrac{m_0 - \mathcal{C}s_0}{1 - \mathcal{C}n}$ which is a positive constant since $\lambda < \tfrac{1}{n-1} \implies \mathcal{C} = \tfrac{\lambda}{1 + \lambda} < \tfrac{1}{n}$. Thus, the fleas' positions are upper bounded by some positive constant, so if we make the starting points far left enough it is impossible to take them all to the right of a given position $M$. $\blacksquare$

Setup We claim that the answer is $\lambda \ge \tfrac{1}{n-1}.$ In the proof, we shall be making use of two variables that change with the flea's position. We define $R$, the position of the right-most flea. Suppose we have fleas $A_1, A_2, \dots, A_n$, and define $d(A_i,A_j)$ to be the positive difference between the positions of flea $A_i$ and flea $A_j.$ Define our second variable, $S$, to be $d(A_1,A_n)+d(A_2,A_n)+\dots + d(A_{n-1},A_n).$ To make our proof easier, we note that if $R\ge M$ at any point in our iteration of moves, we are done since any move that makes fleas jump over this right-most flea will bring that flea past $M$. Bound To prove our bound first, let $\lambda<\tfrac{1}{n-1}$ and we consider a single move, which WLOG starts from a configuration such that $A_1,A_2,\dots, A_n$ are placed in that order from left to right. Then, define $d_i$ to be the initial value of $d(A_i,A_n).$ It is easy to see that $S=d_1+d_2+\dots + d_{n-1}$. Now, suppose $A_i$ jumps over $A_j$, and we split this into several cases. Case 1: $A_i$ ends up to the left of $A_n$ In this case, $S$ decreases and $R$ doesn't increase. We do not elaborate further. Case 2: $A_i$ ends up to the right of $A_n$ In this case, $S$ and $R$ are both maximized when $A_2$ jumps over $A_n$. Indeed, we have $d(A_n,A_i)\le \lambda d_i.$ Thus, $R$ increases by at most $\lambda d_i$ Note that $d(A_j,A_i)\le d(A_j,A_n)+d(A_n,A_i)\le d_j+\lambda d_i$ so \[S\le d_1+\dots + \lambda (n-1)d_i+d_{i+1}+\dots +d_{n-1}\]Thus, $S$ decreases by at least $(1-\lambda (n-1)) d_i.$ Combining both cases, for every $\lambda>0$ we want to increase $R$ by, we must proportionally decrease at least $(1-\lambda (n-1))>0$ from $S$. Since $S$ must be non-negative at all times, $R$ is bounded, so let $M$ be anything that $R$ cannot reach, and our bound is proved. Construction Our construction for $\lambda \ge \tfrac{1}{n-1}$ is as follows: take the left-most flea and jump it across the right-most flea. We again let $A_1,A_2,\dots, A_n$ be placed in that order from left to right and define $d_i$ to be the initial value of $d(A_i,A_n).$ Initially, $S=d_1+d_2+\dots + d_{n-1}.$ After the move, $d(A_i,A_1)$ becomes $d(A_i,A_n)+d(A_n,A_i)=d_i+\lambda d_1$, so $S$ becomes \[\lambda(n-1)d_1+d_2+\dots + d_{n-1}\ge d_1+d_2+\dots + d_{n-1}.\]In particular, this means that the distance between the first and last flea is always at least $\tfrac{S}{n-1}$ so $R$ increases by at least $\tfrac{S}{(n-1)^2}$ each time. Therefore, $R$ is not bounded. We are done.

The answer is all $\lambda \geq \frac{1}{n-1}.$ To prove that this works, we let the leftmost flea jump over the rightmost on each move. If at some moment pairwise distances between consecutive fleas from left to right are $a_1,a_2,\dots ,a_{n-1},$ then after the move they become $a_2,a_3,\dots ,a_{n-1}, \lambda \sum a_i\geq \min a_i$. Therefore the average value of these distances doesn't decrease, so the distance between leftmost and rightmost fleas never decreases too and fleas can jump sufficently far. Now assume $\lambda <\frac{1}{n-1}.$ Numerate fleas by $1,2,\dots ,n$ and let at the moment $x_i$ denotes the coordinate of $i-$th flea. Claim. Number $F=2\max x_i -\sum_{j=1}^n x_j$ is a nonincreasing monovariant under moves. Proof. Let $\varphi$ be a move $x_i\mapsto x_j+\lambda (x_j-x_i)$ for $x_i<x_j.$ Then $$\varphi (F)=\begin{cases} F-\lambda (\varphi(x_i) -x_i)<F,\, &\text{if } \varphi (x_i)<\max x_k \\ F-(1+\lambda)(x_n-x_j)\leq F,\, &\text{otherwise} \quad \Box \end{cases}$$ But for the moment $\max x_k >M$ we obtain $$F>(1-\lambda (n-1))x_n>(1-\lambda (n-1))M\implies M<\frac{\max F}{1-\lambda (n-1)}.$$

very cool

Could not find this invariant until hinted largely ;-; Note that this is equivalent to having one flea travel arbitrarily far. Claim: We can achieve $\lambda \ge \frac{1}{n - 1}$. Proof. Let the fleas have initial starting positions $a_1, a_2, \dots, a_n$. Let $C = \underset{i}{argmin} (a_{i+1} - a_i)$ be the minimum difference between two consecutive fleas. Then $a_n - a_1 \ge (n-1)C$ so jumping $a_1$ over $a_n$ to position $a_{n+1}$ makes it so that $a_{n+1} - a_n \ge C$, so $C$ remains the minimum distance between any two fleas. Repeatedly jumping the last flea over the first flea gives the result. $\blacksquare$ Claim: $\lambda < \frac{1}{n-1}$ is impossible. Proof. Consider the value of $a_n - \lambda (a_1 + \dots + a_{n-1})$ where flies are re-labeled after each turn based on order. We claim that this value is a monovariant that doesn't increase. Unless the farthest fly changes, the value is guarenteed to decrease. Else, suppose that $a_i$ jumps over $a_j$ to become the furthest fly. As $j$ increases, the value afterwards decreases. Thus, assume $j = n$ and that $a_i$ jumps to $(1 + \lambda)a_n - \lambda a_i$. However, then equality holds as \[ (1 + \lambda)a_n - \lambda a_i - \lambda (a_1 + \dots + a_n - a_i) = a_n - \lambda (a_1 + \dots + a_{n-1}). \]Then, it follows that for some $C$ and given initial config \[ C \ge a_n - \lambda (a_1 + \dots + a_{n-1}) \ge a_n (1 - \lambda (n-1)) \]which implies $a_n \le \frac{C}{1 - \lambda(n-1)}$ which bounds growth. $\blacksquare$

This is a very interesting problem. One way to motivate the almost magical invariant is that when you try to further analyze the strategy of making the leftmost flea hop to the rightmost flea is noticing that this is the same as looking at the linear recurrence $x_{n+1} = x_{n} + l(x_n - x_1)$ which has the characteristic polynomial $x^{n+1} = x^{n} + l(x^n - x^1)$ which can be reduced to an $n$ variable one, by noticing that $1$ is a root, dividing gives the expression $x^n - l(x^{n-1} + \cdots + x^1)$, the recurrence $x_m - l(x_{m-1} + \cdots + x_{m-n})$ is now invariant due to construction, but due to $x_1$ symmetry with the other $n-2$ variables terms, then it's also invariant by the others strategies of sending $x_i$ to $x_n$ where $1 \le i \le n-1$.

Clean problem. The answer is $\lambda\geq\frac1{n-1}$. This works by letting the leftmost flea jump over the rightmost flea on each turn, which will never decrease the smallest distance between two fleas, so the process can run forever. Let the flea positions be $p_1\leq p_2\leq\dots\leq p_n$, after every move sort the values of $p$ again. We try to find a monovariant on the positions $p$.

After experimenting on small values of $n$, we get $\frac{p_n-\lambda(p_1+p_2+\dots+p_{n-1})}{1-(n-1)\lambda}$ as a monovariant that works. Any move that keeps $p_n$ the same will only decrease the monovariant, and if something becomes the new furthest it will keep the value the same. For $\lambda<\frac{1}{n-1}$, this the initial value is a positive number which can bound $p_n$ to a finite number because $p_1+p_2+\dots+p_{n-1}$ is bounded below by its initial value, hence we are done.

2000 p3 We claim that $\lambda \geq \frac{1}{n-1}$ Construction: Each jump will be from the leftmost flea over the rightmost one. Since the new distance between them will be the average of the $n-1$ distances, hence the minimum distance between any two fleas will never decrease, proving the construct. Now for the bound. Bound: Let $a_1,a_2 \cdots a_n$ denote the position of the fleas. then we claim that $a_n - \lambda (a_1 + \dots + a_{n-1})$ will eventually be be bounded from above. Proving that we get \begin{align*} a_n - \lambda (a_1 + \dots + a_{n-1}) & >a_n-\lambda(a_n+\cdots+a_n)\\ &>M\big(1-(n-1)\lambda\big), \end{align*} As $M$ approaches infinity this doesn’t hold hence we are done.