{
  "index": "1954-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. Suppose that \\( u_{0}, u_{1}, u_{2}, \\ldots \\) is a sequence of real numbers such that\n\\[\nu_{n}=\\sum_{k=1}^{\\infty} u_{n+k}^{2} \\quad \\text { for } n=0,1,2, \\ldots\n\\]\n\nProve that if \\( \\Sigma u_{n} \\) converges then \\( u_{k}=0 \\) for all \\( k \\).",
  "solution": "Solution. It is obvious from (1) that the terms \\( u_{n} \\) are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using (1) so are all its predecessors.\n\nSuppose \\( \\Sigma u_{n} \\) converges. Let \\( p \\) be chosen so that \\( \\Sigma_{n>p} u_{n}<1 \\). Then, from (1),\n\\[\nu_{p}=\\sum_{n>p} u_{n}^{2} \\leq \\sum_{n>p} u_{p} u_{n}=u_{p} \\sum_{n>p} u_{n} \\leq u_{p}\n\\]\nwith equality at the last step only if \\( u_{p}=0 \\). But, looking at the first member and the last member, we see that equality must hold throughout. Thus \\( u_{r}=0 \\) and hence all the \\( u \\) 's are zero, as we have shown above.",
  "vars": [
    "u_0",
    "u_1",
    "u_2",
    "u_n",
    "u_k",
    "u_r",
    "n",
    "k",
    "r"
  ],
  "params": [
    "u_p",
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "u_0": "zerothterm",
        "u_1": "firstterm",
        "u_2": "secondterm",
        "u_n": "genericterm",
        "u_k": "kthterm",
        "u_r": "rthterm",
        "n": "indexn",
        "k": "indexk",
        "r": "indexr",
        "u_p": "pthterm",
        "p": "indexp"
      },
      "question": "Suppose that \\( zerothterm, firstterm, secondterm, \\ldots \\) is a sequence of real numbers such that\n\\[\ngenericterm=\\sum_{indexk=1}^{\\infty} u_{indexn+indexk}^{2} \\quad \\text { for } indexn=0,1,2, \\ldots\n\\]\n\nProve that if \\( \\Sigma genericterm \\) converges then \\( kthterm=0 \\) for all \\( indexk \\).",
      "solution": "Solution. It is obvious from (1) that the terms \\( genericterm \\) are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using (1) so are all its predecessors.\n\nSuppose \\( \\Sigma genericterm \\) converges. Let \\( indexp \\) be chosen so that \\( \\Sigma_{indexn>indexp} genericterm<1 \\). Then, from (1),\n\\[\npthterm=\\sum_{indexn>indexp} genericterm^{2} \\leq \\sum_{indexn>indexp} pthterm genericterm=pthterm \\sum_{indexn>indexp} genericterm \\leq pthterm\n\\]\nwith equality at the last step only if \\( pthterm=0 \\). But, looking at the first member and the last member, we see that equality must hold throughout. Thus \\( rthterm=0 \\) and hence all the \\( u \\) 's are zero, as we have shown above."
    },
    "descriptive_long_confusing": {
      "map": {
        "u_0": "marigold",
        "u_1": "hazelnuts",
        "u_2": "brickwork",
        "u_n": "buttermilk",
        "u_k": "chandelier",
        "u_r": "dragonfly",
        "n": "sandstone",
        "k": "blueberries",
        "r": "pomegranate",
        "u_p": "paperclips",
        "p": "windstorm"
      },
      "question": "6. Suppose that \\( marigold, hazelnuts, brickwork, \\ldots \\) is a sequence of real numbers such that\n\\[\nbuttermilk=\\sum_{blueberries=1}^{\\infty} u_{sandstone+blueberries}^{2} \\quad \\text { for } sandstone=0,1,2, \\ldots\n\\]\n\nProve that if \\( \\Sigma buttermilk \\) converges then \\( chandelier=0 \\) for all blueberries.",
      "solution": "Solution. It is obvious from (1) that the terms \\( buttermilk \\) are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using (1) so are all its predecessors.\n\nSuppose \\( \\Sigma buttermilk \\) converges. Let \\( windstorm \\) be chosen so that \\( \\Sigma_{sandstone>windstorm} buttermilk<1 \\). Then, from (1),\n\\[\npaperclips=\\sum_{sandstone>windstorm} buttermilk^{2} \\leq \\sum_{sandstone>windstorm} paperclips\\; buttermilk=paperclips \\sum_{sandstone>windstorm} buttermilk \\leq paperclips\n\\]\nwith equality at the last step only if \\( paperclips=0 \\). But, looking at the first member and the last member, we see that equality must hold throughout. Thus \\( dragonfly=0 \\) and hence all the \\( u \\) 's are zero, as we have shown above."
    },
    "descriptive_long_misleading": {
      "map": {
        "u_0": "terminalterm",
        "u_1": "lastentry",
        "u_2": "endingvalue",
        "u_n": "specificindex",
        "u_k": "fixednumber",
        "u_r": "steadycount",
        "n": "immutable",
        "k": "unchanged",
        "r": "fixedunit",
        "u_p": "antiterm",
        "p": "variable"
      },
      "question": "6. Suppose that \\( terminalterm, lastentry, endingvalue, \\ldots \\) is a sequence of real numbers such that\n\\[\nspecificindex=\\sum_{unchanged=1}^{\\infty} u_{immutable+unchanged}^{2} \\quad \\text { for } immutable=0,1,2, \\ldots\n\\]\n\nProve that if \\( \\Sigma specificindex \\) converges then \\( fixednumber=0 \\) for all \\( unchanged \\).",
      "solution": "Solution. It is obvious from (1) that the terms \\( specificindex \\) are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using (1) so are all its predecessors.\n\nSuppose \\( \\Sigma specificindex \\) converges. Let variable be chosen so that \\( \\Sigma_{immutable>variable} u_{immutable}<1 \\). Then, from (1),\n\\[\nantiterm=\\sum_{immutable>variable} specificindex^{2} \\leq \\sum_{immutable>variable} antiterm\\,specificindex = antiterm \\sum_{immutable>variable} specificindex \\leq antiterm\n\\]\nwith equality at the last step only if \\( antiterm=0 \\). But, looking at the first member and the last member, we see that equality must hold throughout. Thus \\( steadycount=0 \\) and hence all the \\( u \\)'s are zero, as we have shown above."
    },
    "garbled_string": {
      "map": {
        "u_0": "xjdkeqra",
        "u_1": "tqwplzbn",
        "u_2": "hrcovjys",
        "u_n": "vgmeulkc",
        "u_k": "pyhzqnof",
        "u_r": "cnwktbfi",
        "n": "dblserxm",
        "k": "zofirnqa",
        "r": "wemlpyut",
        "u_p": "nqidalvh",
        "p": "gskovrje"
      },
      "question": "6. Suppose that \\( xjdkeqra, tqwplzbn, hrcovjys, \\ldots \\) is a sequence of real numbers such that\n\\[\nvgmeulkc=\\sum_{zofirnqa=1}^{\\infty} u_{dblserxm+zofirnqa}^{2} \\quad \\text { for } dblserxm=0,1,2, \\ldots\n\\]\n\nProve that if \\( \\Sigma vgmeulkc \\) converges then \\( pyhzqnof=0 \\) for all \\( zofirnqa \\).",
      "solution": "Solution. It is obvious from (1) that the terms \\( vgmeulkc \\) are non-increasing and non-negative. Thus, if any one term is zero, so are all its successors, and by induction using (1) so are all its predecessors.\n\nSuppose \\( \\Sigma vgmeulkc \\) converges. Let \\( gskovrje \\) be chosen so that \\( \\Sigma_{dblserxm>gskovrje} vgmeulkc<1 \\). Then, from (1),\n\\[\nnqidalvh=\\sum_{dblserxm>gskovrje} vgmeulkc^{2} \\leq \\sum_{dblserxm>gskovrje} nqidalvh vgmeulkc=nqidalvh \\sum_{dblserxm>gskovrje} vgmeulkc \\leq nqidalvh\n\\]\nwith equality at the last step only if \\( nqidalvh=0 \\). But, looking at the first member and the last member, we see that equality must hold throughout. Thus \\( cnwktbfi=0 \\) and hence all the \\( u \\) 's are zero, as we have shown above."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 1$ be a fixed integer.  \nFor every multi–index $k=(k_{1},\\dots ,k_{d})\\in \\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}$ ($\\mathbb{N}^{\\ast}:=\\mathbb{N}\\setminus\\{0\\}$) let $\\omega_{k}$ be a strictly positive real number and put  \n\n\\[\n\\Lambda\\;:=\\;\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k},\n\\qquad 0<\\Lambda<\\infty .\n\\]\n\nA non–negative array $u=\\bigl(u_{n}\\bigr)_{n\\in\\mathbb{N}^{d}}$ is said to satisfy the \\emph{weighted quadratic forward recurrence} if  \n\n\\[\nu_{n}\\;=\\;\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k}\\,u_{\\,n+k}^{\\,2},\n\\qquad \n\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k}\\,u_{\\,n+k}^{\\,2}<\\infty ,\n\\quad n\\in\\mathbb{N}^{d},\n\\]\n\nwhere the sum $n+k$ is taken component–wise.  \n\nAssume in addition that the series of first powers converges:\n\n\\[\n\\sum_{n\\in\\mathbb{N}^{d}}u_{n}<\\infty .\n\\]\n\nProve that necessarily  \n\n\\[\nu_{n}=0\\qquad\\text{for every }n\\in\\mathbb{N}^{d}.\n\\]",
      "solution": "(Throughout, all indices are multi–indices in $\\mathbb{N}^{d}$; boldface is suppressed.  \nThe symbol $\\gtrsim$ means “greater than or equal to up to a harmless algebraic constant”.)\n\n\\medskip\n\\textbf{Step 1 – A quantitative forward lower bound.}  \nFor $n\\in\\mathbb{N}^{d}$ set  \n\n\\[\nM(n):=\\sup_{k\\in\\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}}u_{\\,n+k}\\in[0,\\infty].\n\\]\n\nBecause $\\omega_{k}>0$ and $\\Lambda<\\infty$ we obtain for every $n$  \n\n\\[\nu_{n}\\;=\\;\\sum_{k}\\omega_{k}u_{\\,n+k}^{2}\\;\\le\\;\n\\Lambda\\,M(n)^{2}.\n\\]\n\nHence, whenever $u_{n}>0$ we have  \n\n\\[\nM(n)\\;\\ge\\;\\sqrt{\\frac{u_{n}}{\\Lambda}}. \\tag{1}\n\\]\n\n\\medskip\n\\textbf{Step 2 – Constructing an infinitely shifted chain.}  \nAssume, for a contradiction, that $u$ is \\emph{not} identically zero and choose an index $n(0)$ with $u_{\\,n(0)}>0$.\n\n\\smallskip\n\\emph{Lemma (single–step growth).}  \nFor every $n$ with $u_{n}>0$ there exists $k(n)\\in\\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}$ such that  \n\n\\[\nu_{\\,n+k(n)}\\;\\ge\\;\\sqrt{\\frac{u_{n}}{\\Lambda}}. \\tag{2}\n\\]\n\n\\emph{Proof.}  \nIf $u_{\\,n+k}<\\sqrt{u_{n}/\\Lambda}$ for all $k$, then  \n\n\\[\nu_{n}\n=\\sum_{k}\\omega_{k}u_{\\,n+k}^{2}\n<\\sum_{k}\\omega_{k}\\Bigl(\\frac{u_{n}}{\\Lambda}\\Bigr)\n=\\frac{u_{n}}{\\Lambda}\\,\\Lambda=u_{n},\n\\]\n\na contradiction. \\hfill $\\square$\n\n\\smallskip\nDefine recursively  \n\n\\[\nn(\\ell+1):=n(\\ell)+k\\bigl(n(\\ell)\\bigr),\n\\qquad \\ell=0,1,2,\\dots .\n\\]\n\nAll $d$ coordinates of $n(\\ell)$ strictly increase with $\\ell$, so the indices $n(\\ell)$ are pairwise distinct.  \nSet  \n\n\\[\nx_{\\ell}:=u_{\\,n(\\ell)},\\qquad \\ell=0,1,2,\\dots .\n\\]\n\nBy the lemma,\n\n\\[\nx_{\\ell+1}\\;\\ge\\;\\sqrt{\\frac{x_{\\ell}}{\\Lambda}}\\qquad(\\ell\\ge 0). \\tag{3}\n\\]\n\n\\medskip\n\\textbf{Step 3 – Explicit control of the iterates and a uniform positive lower bound.}  \nIntroduce the map  \n\n\\[\n\\varphi(t):=\\sqrt{\\frac{t}{\\Lambda}},\\qquad t\\ge 0.\n\\]\n\nBecause $\\varphi$ is increasing, inequality (3) yields by induction  \n\n\\[\nx_{\\ell}\\;\\ge\\;\\varphi^{(\\ell)}(x_{0}),\n\\qquad \\ell=0,1,2,\\dots ,\n\\]\n\nwhere $\\varphi^{(\\ell)}$ denotes the $\\ell$–fold iterate of $\\varphi$.  \nA direct computation shows  \n\n\\[\n\\varphi^{(\\ell)}(t)=\\Lambda^{-1}\\bigl(\\Lambda t\\bigr)^{2^{-\\ell}},\\qquad t\\ge 0.\n\\]\n\nConsequently\n\n\\[\nx_{\\ell}\\;\\ge\\;\\Lambda^{-1}\\bigl(\\Lambda x_{0}\\bigr)^{2^{-\\ell}}. \\tag{4}\n\\]\n\nSince $0<(\\Lambda x_{0})^{2^{-\\ell}}\\longrightarrow 1$ as $\\ell\\to\\infty$, relation (4) implies  \n\n\\[\n\\liminf_{\\ell\\to\\infty}x_{\\ell}\\;\\ge\\;\\Lambda^{-1}. \\tag{5}\n\\]\n\n(Note that we do \\emph{not} assert the existence of $\\displaystyle\\lim_{\\ell\\to\\infty}x_{\\ell}$; only the lower bound (5) is needed.)\n\nChoose the constant  \n\n\\[\n\\delta:=\\frac{1}{2}\\,\\Lambda^{-1}\\quad\\bigl(0<\\delta<\\Lambda^{-1}\\bigr).\n\\]\n\nBy (5) there exists $\\ell_{0}\\in\\mathbb{N}$ such that  \n\n\\[\nx_{\\ell}\\;\\ge\\;\\delta\\qquad\\text{for every }\\ell\\ge\\ell_{0}. \\tag{6}\n\\]\n\n\\medskip\n\\textbf{Step 4 – Divergence of the total mass.}  \nBecause the indices $n(\\ell)$ are distinct, the convergent series in the hypothesis contains the subseries  \n\n\\[\n\\sum_{\\ell\\ge\\ell_{0}}u_{\\,n(\\ell)}\n=\\sum_{\\ell\\ge\\ell_{0}}x_{\\ell}.\n\\]\n\nBy (6) every term of this subseries is at least $\\delta>0$, whence its partial sums satisfy  \n\n\\[\nS_{N}:=\\sum_{\\ell=\\ell_{0}}^{\\ell_{0}+N}x_{\\ell}\n\\;\\ge\\;(N+1)\\,\\delta\\;\\longrightarrow\\;\\infty\\quad(N\\to\\infty),\n\\]\n\ncontradicting the assumed convergence of $\\sum_{n}u_{n}$.\n\n\\medskip\n\\textbf{Step 5 – Conclusion.}  \nThe contradiction shows that no positive entry can exist.  \nBecause $u$ is non–negative, it follows that $u_{n}=0$ for every $n\\in\\mathbb{N}^{d}$.  \\hfill $\\square$\n\n\\medskip\n\\emph{Remark.}  \nThe above argument demonstrates not only that $u$ must vanish identically, but also that any non–negative solution of the weighted quadratic recurrence whose sum of first powers diverges is forced to have a subsequence bounded \\emph{below} by a positive constant determined solely by $\\Lambda$.  No comparable \\emph{upper} bound can be deduced in general, showing that the lower–bound technique employed here is essentially sharp.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.468564",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  The problem passes from a single sequence (one-dimensional index) to an array indexed by ℕ^{d}.  The partial order and the “future cone’’ Γ_n necessitate multi-index notation and more elaborate induction.\n\n2. Additional structure:  A non-trivial weight system (ω_k) with only a finiteness condition has been introduced; the argument must work uniformly for arbitrary positive weights whose total mass is merely finite.\n\n3. Technical subtleties:  \n   •  Proving monotonicity now uses set inclusion of cones rather than a simple shift.  \n   •  Tail estimates require multi-dimensional summation and careful control via Λ.  \n   •  The vanishing of all indices is achieved through a reverse-level induction on the sum of coordinates, a technique absent from the original exercise.\n\n4. Deeper insight:  One has to realise that the key inequality u_{N+k}^{2} ≤ u_N u_{N+k} follows from monotonicity in d coordinates and that the parameter Λ allows a quantitative bound strong enough to force u_N to zero.\n\n5. Larger solution path:  Compared with the original two-line contradiction, the enhanced variant demands five conceptual steps (monotonicity, tail control, weighted inequality, annihilation of the cone, downward induction), each relying on multi-index combinatorics and weighted estimates.\n\nConsequently the enhanced kernel variant is significantly more intricate and requires a broader toolkit than the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 1$ be a fixed integer.  \nFor every multi–index $k=(k_{1},\\dots ,k_{d})\\in \\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}$ ($\\mathbb{N}^{\\ast}:=\\mathbb{N}\\setminus\\{0\\}$) let $\\omega_{k}$ be a strictly positive real number and put  \n\n\\[\n\\Lambda\\;:=\\;\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k},\n\\qquad 0<\\Lambda<\\infty .\n\\]\n\nA non–negative array $u=\\bigl(u_{n}\\bigr)_{n\\in\\mathbb{N}^{d}}$ is said to satisfy the \\emph{weighted quadratic forward recurrence} if  \n\n\\[\nu_{n}\\;=\\;\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k}\\,u_{\\,n+k}^{\\,2},\n\\qquad \n\\sum_{k\\in\\left(\\mathbb{N}^{\\ast}\\right)^{d}}\\omega_{k}\\,u_{\\,n+k}^{\\,2}<\\infty ,\n\\quad n\\in\\mathbb{N}^{d},\n\\]\n\nwhere the sum $n+k$ is taken component–wise.  \n\nAssume in addition that the series of first powers converges:\n\n\\[\n\\sum_{n\\in\\mathbb{N}^{d}}u_{n}<\\infty .\n\\]\n\nProve that necessarily  \n\n\\[\nu_{n}=0\\qquad\\text{for every }n\\in\\mathbb{N}^{d}.\n\\]",
      "solution": "(Throughout, all indices are multi–indices in $\\mathbb{N}^{d}$; boldface is suppressed.  \nThe symbol $\\gtrsim$ means “greater than or equal to up to a harmless algebraic constant”.)\n\n\\medskip\n\\textbf{Step 1 – A quantitative forward lower bound.}  \nFor $n\\in\\mathbb{N}^{d}$ set  \n\n\\[\nM(n):=\\sup_{k\\in\\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}}u_{\\,n+k}\\in[0,\\infty].\n\\]\n\nBecause $\\omega_{k}>0$ and $\\Lambda<\\infty$ we obtain for every $n$  \n\n\\[\nu_{n}\\;=\\;\\sum_{k}\\omega_{k}u_{\\,n+k}^{2}\\;\\le\\;\n\\Lambda\\,M(n)^{2}.\n\\]\n\nHence, whenever $u_{n}>0$ we have  \n\n\\[\nM(n)\\;\\ge\\;\\sqrt{\\frac{u_{n}}{\\Lambda}}. \\tag{1}\n\\]\n\n\\medskip\n\\textbf{Step 2 – Constructing an infinitely shifted chain.}  \nAssume, for a contradiction, that $u$ is \\emph{not} identically zero and choose an index $n(0)$ with $u_{\\,n(0)}>0$.\n\n\\smallskip\n\\emph{Lemma (single–step growth).}  \nFor every $n$ with $u_{n}>0$ there exists $k(n)\\in\\bigl(\\mathbb{N}^{\\ast}\\bigr)^{d}$ such that  \n\n\\[\nu_{\\,n+k(n)}\\;\\ge\\;\\sqrt{\\frac{u_{n}}{\\Lambda}}. \\tag{2}\n\\]\n\n\\emph{Proof.}  \nIf $u_{\\,n+k}<\\sqrt{u_{n}/\\Lambda}$ for all $k$, then  \n\n\\[\nu_{n}\n=\\sum_{k}\\omega_{k}u_{\\,n+k}^{2}\n<\\sum_{k}\\omega_{k}\\Bigl(\\frac{u_{n}}{\\Lambda}\\Bigr)\n=\\frac{u_{n}}{\\Lambda}\\,\\Lambda=u_{n},\n\\]\n\na contradiction. \\hfill $\\square$\n\n\\smallskip\nDefine recursively  \n\n\\[\nn(\\ell+1):=n(\\ell)+k\\bigl(n(\\ell)\\bigr),\n\\qquad \\ell=0,1,2,\\dots .\n\\]\n\nAll $d$ coordinates of $n(\\ell)$ strictly increase with $\\ell$, so the indices $n(\\ell)$ are pairwise distinct.  \nSet  \n\n\\[\nx_{\\ell}:=u_{\\,n(\\ell)},\\qquad \\ell=0,1,2,\\dots .\n\\]\n\nBy the lemma,\n\n\\[\nx_{\\ell+1}\\;\\ge\\;\\sqrt{\\frac{x_{\\ell}}{\\Lambda}}\\qquad(\\ell\\ge 0). \\tag{3}\n\\]\n\n\\medskip\n\\textbf{Step 3 – Explicit control of the iterates and a uniform positive lower bound.}  \nIntroduce the map  \n\n\\[\n\\varphi(t):=\\sqrt{\\frac{t}{\\Lambda}},\\qquad t\\ge 0.\n\\]\n\nBecause $\\varphi$ is increasing, inequality (3) yields by induction  \n\n\\[\nx_{\\ell}\\;\\ge\\;\\varphi^{(\\ell)}(x_{0}),\n\\qquad \\ell=0,1,2,\\dots ,\n\\]\n\nwhere $\\varphi^{(\\ell)}$ denotes the $\\ell$–fold iterate of $\\varphi$.  \nA direct computation shows  \n\n\\[\n\\varphi^{(\\ell)}(t)=\\Lambda^{-1}\\bigl(\\Lambda t\\bigr)^{2^{-\\ell}},\\qquad t\\ge 0.\n\\]\n\nConsequently\n\n\\[\nx_{\\ell}\\;\\ge\\;\\Lambda^{-1}\\bigl(\\Lambda x_{0}\\bigr)^{2^{-\\ell}}. \\tag{4}\n\\]\n\nSince $0<(\\Lambda x_{0})^{2^{-\\ell}}\\longrightarrow 1$ as $\\ell\\to\\infty$, relation (4) implies  \n\n\\[\n\\liminf_{\\ell\\to\\infty}x_{\\ell}\\;\\ge\\;\\Lambda^{-1}. \\tag{5}\n\\]\n\n(Note that we do \\emph{not} assert the existence of $\\displaystyle\\lim_{\\ell\\to\\infty}x_{\\ell}$; only the lower bound (5) is needed.)\n\nChoose the constant  \n\n\\[\n\\delta:=\\frac{1}{2}\\,\\Lambda^{-1}\\quad\\bigl(0<\\delta<\\Lambda^{-1}\\bigr).\n\\]\n\nBy (5) there exists $\\ell_{0}\\in\\mathbb{N}$ such that  \n\n\\[\nx_{\\ell}\\;\\ge\\;\\delta\\qquad\\text{for every }\\ell\\ge\\ell_{0}. \\tag{6}\n\\]\n\n\\medskip\n\\textbf{Step 4 – Divergence of the total mass.}  \nBecause the indices $n(\\ell)$ are distinct, the convergent series in the hypothesis contains the subseries  \n\n\\[\n\\sum_{\\ell\\ge\\ell_{0}}u_{\\,n(\\ell)}\n=\\sum_{\\ell\\ge\\ell_{0}}x_{\\ell}.\n\\]\n\nBy (6) every term of this subseries is at least $\\delta>0$, whence its partial sums satisfy  \n\n\\[\nS_{N}:=\\sum_{\\ell=\\ell_{0}}^{\\ell_{0}+N}x_{\\ell}\n\\;\\ge\\;(N+1)\\,\\delta\\;\\longrightarrow\\;\\infty\\quad(N\\to\\infty),\n\\]\n\ncontradicting the assumed convergence of $\\sum_{n}u_{n}$.\n\n\\medskip\n\\textbf{Step 5 – Conclusion.}  \nThe contradiction shows that no positive entry can exist.  \nBecause $u$ is non–negative, it follows that $u_{n}=0$ for every $n\\in\\mathbb{N}^{d}$.  \\hfill $\\square$\n\n\\medskip\n\\emph{Remark.}  \nThe above argument demonstrates not only that $u$ must vanish identically, but also that any non–negative solution of the weighted quadratic recurrence whose sum of first powers diverges is forced to have a subsequence bounded \\emph{below} by a positive constant determined solely by $\\Lambda$.  No comparable \\emph{upper} bound can be deduced in general, showing that the lower–bound technique employed here is essentially sharp.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.393744",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  The problem passes from a single sequence (one-dimensional index) to an array indexed by ℕ^{d}.  The partial order and the “future cone’’ Γ_n necessitate multi-index notation and more elaborate induction.\n\n2. Additional structure:  A non-trivial weight system (ω_k) with only a finiteness condition has been introduced; the argument must work uniformly for arbitrary positive weights whose total mass is merely finite.\n\n3. Technical subtleties:  \n   •  Proving monotonicity now uses set inclusion of cones rather than a simple shift.  \n   •  Tail estimates require multi-dimensional summation and careful control via Λ.  \n   •  The vanishing of all indices is achieved through a reverse-level induction on the sum of coordinates, a technique absent from the original exercise.\n\n4. Deeper insight:  One has to realise that the key inequality u_{N+k}^{2} ≤ u_N u_{N+k} follows from monotonicity in d coordinates and that the parameter Λ allows a quantitative bound strong enough to force u_N to zero.\n\n5. Larger solution path:  Compared with the original two-line contradiction, the enhanced variant demands five conceptual steps (monotonicity, tail control, weighted inequality, annihilation of the cone, downward induction), each relying on multi-index combinatorics and weighted estimates.\n\nConsequently the enhanced kernel variant is significantly more intricate and requires a broader toolkit than the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}