{
  "index": "2001-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "For any positive integer $n$, let $\\langle n\\rangle$ denote\nthe closest integer to $\\sqrt{n}$.  Evaluate\n\\[\\sum_{n=1}^\\infty \\frac{2^{\\langle n\\rangle}+2^{-\\langle n\\rangle}}\n                         {2^n}.\\]",
  "solution": "Since $(k-1/2)^2 = k^2-k+1/4$ and $(k+1/2)^2 = k^2+k+1/4$,\nwe have that $\\langle n \\rangle = k$ if and only if\n$k^2-k+1 \\leq n \\leq k^2+k$.  Hence\n\\begin{align*}\n\\sum_{n=1}^\\infty \\frac{2^{\\langle n \\rangle} + 2^{-\\langle n \\rangle}}{2^n}\n&= \\sum_{k=1}^\\infty \\sum_{n, \\langle n \\rangle = k}\n    \\frac{2^{\\langle n \\rangle} + 2^{-\\langle n \\rangle}}{2^n} \\\\\n&= \\sum_{k=1}^\\infty \\sum_{n=k^2-k+1}^{k^2+k} \\frac{2^k+2^{-k}}{2^n} \\\\\n&= \\sum_{k=1}^\\infty (2^k+2^{-k})(2^{-k^2+k}-2^{-k^2-k}) \\\\\n&= \\sum_{k=1}^\\infty (2^{-k(k-2)} - 2^{-k(k+2)}) \\\\\n&= \\sum_{k=1}^\\infty 2^{-k(k-2)} - \\sum_{k=3}^\\infty 2^{-k(k-2)} \\\\\n&= 3.\n\\end{align*}\n\nAlternate solution: rewrite the sum as $\\sum_{n=1}^\\infty\n2^{-(n+\\langle n \\rangle)} + \\sum_{n=1}^\\infty\n2^{-(n - \\langle n \\rangle)}$.\nNote that $\\langle n \\rangle \\neq \\langle n+1 \\rangle$\nif and only if $n = m^2+m$ for some $m$. Thus $n + \\langle n \\rangle$\nand $n - \\langle n \\rangle$ each increase by 1 except at $n=m^2+m$,\nwhere the former skips from $m^2+2m$ to $m^2+2m+2$ and the latter\nrepeats the value $m^2$. Thus the sums are\n\\[\n\\sum_{n=1}^\\infty 2^{-n} - \\sum_{m=1}^\\infty 2^{-m^2}\n+ \\sum_{n=0}^\\infty 2^{-n} + \\sum_{m=1}^\\infty 2^{-m^2}\n= 2+1=3.\n\\]",
  "vars": [
    "n",
    "k",
    "m"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexn",
        "k": "loopkk",
        "m": "templm"
      },
      "question": "For any positive integer $indexn$, let $\\langle indexn\\rangle$ denote\nthe closest integer to $\\sqrt{indexn}$.  Evaluate\n\\[\\sum_{indexn=1}^\\infty \\frac{2^{\\langle indexn\\rangle}+2^{-\\langle indexn\\rangle}}\n                         {2^{indexn}}.\\]",
      "solution": "Since $(loopkk-1/2)^2 = loopkk^2-loopkk+1/4$ and $(loopkk+1/2)^2 = loopkk^2+loopkk+1/4$,\nwe have that $\\langle indexn \\rangle = loopkk$ if and only if\n$loopkk^2-loopkk+1 \\leq indexn \\leq loopkk^2+loopkk$.  Hence\n\\begin{align*}\n\\sum_{indexn=1}^\\infty \\frac{2^{\\langle indexn \\rangle} + 2^{-\\langle indexn \\rangle}}{2^{indexn}}\n&= \\sum_{loopkk=1}^\\infty \\sum_{indexn, \\langle indexn \\rangle = loopkk}\n    \\frac{2^{\\langle indexn \\rangle} + 2^{-\\langle indexn \\rangle}}{2^{indexn}} \\\\\n&= \\sum_{loopkk=1}^\\infty \\sum_{indexn=loopkk^2-loopkk+1}^{loopkk^2+loopkk} \\frac{2^{loopkk}+2^{-loopkk}}{2^{indexn}} \\\\\n&= \\sum_{loopkk=1}^\\infty (2^{loopkk}+2^{-loopkk})(2^{-loopkk^2+loopkk}-2^{-loopkk^2-loopkk}) \\\\\n&= \\sum_{loopkk=1}^\\infty (2^{-loopkk(loopkk-2)} - 2^{-loopkk(loopkk+2)}) \\\\\n&= \\sum_{loopkk=1}^\\infty 2^{-loopkk(loopkk-2)} - \\sum_{loopkk=3}^\\infty 2^{-loopkk(loopkk-2)} \\\\\n&= 3.\n\\end{align*}\n\nAlternate solution: rewrite the sum as $\\sum_{indexn=1}^\\infty\n2^{-(indexn+\\langle indexn \\rangle)} + \\sum_{indexn=1}^\\infty\n2^{-(indexn - \\langle indexn \\rangle)}$.\nNote that $\\langle indexn \\rangle \\neq \\langle indexn+1 \\rangle$\nif and only if $indexn = templm^2+templm$ for some $templm$. Thus $indexn + \\langle indexn \\rangle$\nand $indexn - \\langle indexn \\rangle$ each increase by 1 except at $indexn=templm^2+templm$,\nwhere the former skips from $templm^2+2templm$ to $templm^2+2templm+2$ and the latter\nrepeats the value $templm^2$. Thus the sums are\n\\[\n\\sum_{indexn=1}^\\infty 2^{-indexn} - \\sum_{templm=1}^\\infty 2^{-templm^2}\n+ \\sum_{indexn=0}^\\infty 2^{-indexn} + \\sum_{templm=1}^\\infty 2^{-templm^2}\n= 2+1=3.\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "sunflower",
        "k": "blueberry",
        "m": "calendar"
      },
      "question": "For any positive integer sunflower, let $\\langle sunflower\\rangle$ denote\nthe closest integer to $\\sqrt{sunflower}$.  Evaluate\n\\[\\sum_{sunflower=1}^\\infty \\frac{2^{\\langle sunflower\\rangle}+2^{-\\langle sunflower\\rangle}}\n                         {2^{sunflower}}.\\]",
      "solution": "Since $(blueberry-1/2)^2 = blueberry^2-blueberry+1/4$ and $(blueberry+1/2)^2 = blueberry^2+blueberry+1/4$,\nwe have that $\\langle sunflower \\rangle = blueberry$ if and only if\n$blueberry^2-blueberry+1 \\leq sunflower \\leq blueberry^2+blueberry$.  Hence\n\\begin{align*}\n\\sum_{sunflower=1}^\\infty \\frac{2^{\\langle sunflower \\rangle} + 2^{-\\langle sunflower \\rangle}}{2^{sunflower}}\n&= \\sum_{blueberry=1}^\\infty \\sum_{sunflower, \\langle sunflower \\rangle = blueberry}\n    \\frac{2^{\\langle sunflower \\rangle} + 2^{-\\langle sunflower \\rangle}}{2^{sunflower}} \\\\\n&= \\sum_{blueberry=1}^\\infty \\sum_{sunflower=blueberry^2-blueberry+1}^{blueberry^2+blueberry} \\frac{2^{blueberry}+2^{-blueberry}}{2^{sunflower}} \\\\\n&= \\sum_{blueberry=1}^\\infty (2^{blueberry}+2^{-blueberry})(2^{-blueberry^2+blueberry}-2^{-blueberry^2-blueberry}) \\\\\n&= \\sum_{blueberry=1}^\\infty (2^{-blueberry(blueberry-2)} - 2^{-blueberry(blueberry+2)}) \\\\\n&= \\sum_{blueberry=1}^\\infty 2^{-blueberry(blueberry-2)} - \\sum_{blueberry=3}^\\infty 2^{-blueberry(blueberry-2)} \\\\\n&= 3.\n\\end{align*}\n\nAlternate solution: rewrite the sum as $\\sum_{sunflower=1}^\\infty\n2^{-(sunflower+\\langle sunflower \\rangle)} + \\sum_{sunflower=1}^\\infty\n2^{-(sunflower - \\langle sunflower \\rangle)}$.\nNote that $\\langle sunflower \\rangle \\neq \\langle sunflower+1 \\rangle$\nif and only if $sunflower = calendar^2+calendar$ for some $calendar$. Thus $sunflower + \\langle sunflower \\rangle$\nand $sunflower - \\langle sunflower \\rangle$ each increase by 1 except at $sunflower=calendar^2+calendar$,\nwhere the former skips from $calendar^2+2calendar$ to $calendar^2+2calendar+2$ and the latter\nrepeats the value $calendar^2$. Thus the sums are\n\\[\n\\sum_{sunflower=1}^\\infty 2^{-sunflower} - \\sum_{calendar=1}^\\infty 2^{-calendar^2}\n+ \\sum_{sunflower=0}^\\infty 2^{-sunflower} + \\sum_{calendar=1}^\\infty 2^{-calendar^2}\n= 2+1=3.\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "noninteger",
        "k": "irrationalvalue",
        "m": "fractional"
      },
      "question": "For any positive integer $noninteger$, let $\\langle noninteger\\rangle$ denote\nthe closest integer to $\\sqrt{noninteger}$.  Evaluate\n\\[\\sum_{noninteger=1}^\\infty \\frac{2^{\\langle noninteger\\rangle}+2^{-\\langle noninteger\\rangle}}\n                         {2^{noninteger}}.\\]",
      "solution": "Since $(irrationalvalue-1/2)^2 = irrationalvalue^2-irrationalvalue+1/4$ and $(irrationalvalue+1/2)^2 = irrationalvalue^2+irrationalvalue+1/4$,\nwe have that $\\langle noninteger \\rangle = irrationalvalue$ if and only if\n$irrationalvalue^2-irrationalvalue+1 \\leq noninteger \\leq irrationalvalue^2+irrationalvalue$.  Hence\n\\begin{align*}\n\\sum_{noninteger=1}^\\infty \\frac{2^{\\langle noninteger \\rangle} + 2^{-\\langle noninteger \\rangle}}{2^{noninteger}}\n&= \\sum_{irrationalvalue=1}^\\infty \\sum_{noninteger, \\langle noninteger \\rangle = irrationalvalue}\n    \\frac{2^{\\langle noninteger \\rangle} + 2^{-\\langle noninteger \\rangle}}{2^{noninteger}} \\\\\n&= \\sum_{irrationalvalue=1}^\\infty \\sum_{noninteger=irrationalvalue^2-irrationalvalue+1}^{irrationalvalue^2+irrationalvalue} \\frac{2^{irrationalvalue}+2^{-irrationalvalue}}{2^{noninteger}} \\\\\n&= \\sum_{irrationalvalue=1}^\\infty (2^{irrationalvalue}+2^{-irrationalvalue})(2^{-irrationalvalue^2+irrationalvalue}-2^{-irrationalvalue^2-irrationalvalue}) \\\\\n&= \\sum_{irrationalvalue=1}^\\infty (2^{-irrationalvalue(irrationalvalue-2)} - 2^{-irrationalvalue(irrationalvalue+2)}) \\\\\n&= \\sum_{irrationalvalue=1}^\\infty 2^{-irrationalvalue(irrationalvalue-2)} - \\sum_{irrationalvalue=3}^\\infty 2^{-irrationalvalue(irrationalvalue-2)} \\\\\n&= 3.\n\\end{align*}\n\nAlternate solution: rewrite the sum as $\\sum_{noninteger=1}^\\infty\n2^{-(noninteger+\\langle noninteger \\rangle)} + \\sum_{noninteger=1}^\\infty\n2^{-(noninteger - \\langle noninteger \\rangle)}$.\nNote that $\\langle noninteger \\rangle \\neq \\langle noninteger+1 \\rangle$\nif and only if $noninteger = fractional^2+fractional$ for some $fractional$. Thus $noninteger + \\langle noninteger \\rangle$\nand $noninteger - \\langle noninteger \\rangle$ each increase by 1 except at $noninteger=fractional^2+fractional$,\nwhere the former skips from $fractional^2+2fractional$ to $fractional^2+2fractional+2$ and the latter\nrepeats the value $fractional^2$. Thus the sums are\n\\[\n\\sum_{noninteger=1}^\\infty 2^{-noninteger} - \\sum_{fractional=1}^\\infty 2^{-fractional^2}\n+ \\sum_{noninteger=0}^\\infty 2^{-noninteger} + \\sum_{fractional=1}^\\infty 2^{-fractional^2}\n= 2+1=3.\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "k": "hjgrksla",
        "m": "pdrvqble"
      },
      "question": "For any positive integer $qzxwvtnp$, let $\\langle qzxwvtnp\\rangle$ denote\nthe closest integer to $\\sqrt{qzxwvtnp}$.  Evaluate\n\\[\\sum_{qzxwvtnp=1}^\\infty \\frac{2^{\\langle qzxwvtnp\\rangle}+2^{-\\langle qzxwvtnp\\rangle}}\n                         {2^{qzxwvtnp}}.\\]",
      "solution": "Since $(hjgrksla-1/2)^2 = hjgrksla^2-hjgrksla+1/4$ and $(hjgrksla+1/2)^2 = hjgrksla^2+hjgrksla+1/4$,\nwe have that $\\langle qzxwvtnp \\rangle = hjgrksla$ if and only if\n$hjgrksla^2-hjgrksla+1 \\le qzxwvtnp \\le hjgrksla^2+hjgrksla$.  Hence\n\\begin{align*}\n\\sum_{qzxwvtnp=1}^\\infty \\frac{2^{\\langle qzxwvtnp \\rangle} + 2^{-\\langle qzxwvtnp \\rangle}}{2^{qzxwvtnp}}\n&= \\sum_{hjgrksla=1}^\\infty \\sum_{qzxwvtnp, \\langle qzxwvtnp \\rangle = hjgrksla}\n    \\frac{2^{\\langle qzxwvtnp \\rangle} + 2^{-\\langle qzxwvtnp \\rangle}}{2^{qzxwvtnp}} \\\\\n&= \\sum_{hjgrksla=1}^\\infty \\sum_{qzxwvtnp=hjgrksla^2-hjgrksla+1}^{hjgrksla^2+hjgrksla} \\frac{2^{hjgrksla}+2^{-hjgrksla}}{2^{qzxwvtnp}} \\\\\n&= \\sum_{hjgrksla=1}^\\infty (2^{hjgrksla}+2^{-hjgrksla})(2^{-hjgrksla^2+hjgrksla}-2^{-hjgrksla^2-hjgrksla}) \\\\\n&= \\sum_{hjgrksla=1}^\\infty (2^{-hjgrksla(hjgrksla-2)} - 2^{-hjgrksla(hjgrksla+2)}) \\\\\n&= \\sum_{hjgrksla=1}^\\infty 2^{-hjgrksla(hjgrksla-2)} - \\sum_{hjgrksla=3}^\\infty 2^{-hjgrksla(hjgrksla-2)} \\\\\n&= 3.\n\\end{align*}\n\nAlternate solution: rewrite the sum as $\\sum_{qzxwvtnp=1}^\\infty\n2^{-(qzxwvtnp+\\langle qzxwvtnp \\rangle)} + \\sum_{qzxwvtnp=1}^\\infty\n2^{-(qzxwvtnp - \\langle qzxwvtnp \\rangle)}$.\nNote that $\\langle qzxwvtnp \\rangle \\neq \\langle qzxwvtnp+1 \\rangle$\nif and only if $qzxwvtnp = pdrvqble^2+pdrvqble$ for some $pdrvqble$. Thus $qzxwvtnp + \\langle qzxwvtnp \\rangle$\nand $qzxwvtnp - \\langle qzxwvtnp \\rangle$ each increase by 1 except at $qzxwvtnp=pdrvqble^2+pdrvqble$,\nwhere the former skips from $pdrvqble^2+2pdrvqble$ to $pdrvqble^2+2pdrvqble+2$ and the latter\nrepeats the value $pdrvqble^2$. Thus the sums are\n\\[\n\\sum_{qzxwvtnp=1}^\\infty 2^{-qzxwvtnp} - \\sum_{pdrvqble=1}^\\infty 2^{-pdrvqble^2}\n+ \\sum_{qzxwvtnp=0}^\\infty 2^{-qzxwvtnp} + \\sum_{pdrvqble=1}^\\infty 2^{-pdrvqble^2}\n= 2+1=3.\n\\]"
    },
    "kernel_variant": {
      "question": "Fix a real parameter $b>1$.  \nFor every positive integer $n$ let $\\langle n\\rangle$ denote the integer that is closest to $\\sqrt n$ (if $\\sqrt n$ is half–integer, either neighbour may be chosen).  \nProve that the alternating series  \n\n\\[\nS(b)=\\sum_{n=1}^{\\infty}(-1)^{n}\\,\n      \\frac{\\,b^{\\langle n\\rangle}+b^{-\\langle n\\rangle}\\!}{b^{\\,n}}\n\\]\n\nconverges for every such $b$, and show that its sum is in fact  \n\\[\nS(b)=-1,\n\\qquad\\text{independent of }b.\n\\]",
      "solution": "Step 1 –  A convenient parametrisation of the set $\\{n\\ge 1:\\langle n\\rangle=k\\}$.  \nFor each positive integer $k$,\n\\[\n\\bigl(k-\\tfrac12\\bigr)^{\\!2}\\le n<\\bigl(k+\\tfrac12\\bigr)^{\\!2}\n\\Longleftrightarrow\nk^{2}-k+1\\;\\le\\; n\\;\\le\\; k^{2}+k .\n\\]\nHence\n\\[\nI_k:=\\{n:\\langle n\\rangle=k\\}\n      =\\{k^{2}-k+1,\\;k^{2}-k+2,\\ldots ,k^{2}+k\\},\n\\]\nso $|I_k|=2k$ and the smallest index in $I_k$ is\n\\(\na_k=k^{2}-k+1.\n\\)\n\nStep 2 –  Parity of the first index in $I_k$.  \nBecause $k^{2}-k$ is the product of consecutive integers, it is even; hence  \n\\(\na_k=k^{2}-k+1\n\\)\nis always odd.  Consequently\n\\(\n(-1)^{a_k}=-1\n\\)\nfor every $k$.\n\nStep 3 –  Writing $S(b)$ as a sum over the blocks $I_k$.  \nWithin block $I_k$ the factor $b^{\\langle n\\rangle}+b^{-\\langle n\\rangle}$\nis constant and equal to $b^k+b^{-k}$.  Thus\n\\[\nS(b)=\\sum_{k=1}^{\\infty}(b^k+b^{-k})\n      \\sum_{n\\in I_k}(-1)^{n}\\,b^{-n}.\n\\]\n\nStep 4 –  Evaluating the inner geometric sum.  \nPut $q:=-\\dfrac1b$ so that $|q|<1$ and\n\\(\n(-1)^n b^{-n}=q^{\\,n}.\n\\)\nFor $n=a_k+i\\;(0\\le i<2k)$ we have\n\\[\nq^{\\,n}=q^{\\,a_k+i}=q^{\\,a_k}\\,q^{\\,i}.\n\\]\nBecause $q^{\\,a_k}=(-1)^{a_k}b^{-a_k}=-\\,b^{-a_k}$ by Step 2,\n\\[\n\\sum_{n\\in I_k}(-1)^{n}b^{-n}\n     =-\\,b^{-a_k}\\sum_{i=0}^{2k-1}q^{\\,i}\n     =-\\,b^{-a_k}\\,\\frac{1-q^{2k}}{1-q},\n\\]\nand with $q=-\\tfrac1b$ this becomes\n\\[\n\\sum_{n\\in I_k}(-1)^{n}b^{-n}\n     =-\\frac{b^{-a_k}\\bigl(1-b^{-2k}\\bigr)}{1+1/b}\n     =-\\frac{b}{b+1}\\,b^{-a_k}\\bigl(1-b^{-2k}\\bigr).\n\\]\n\nStep 5 –  A crucial algebraic factorisation.  \nNote that\n\\[\n(b^k+b^{-k})\\bigl(1-b^{-2k}\\bigr)=b^k-b^{-3k}.\n\\]\nHence the total contribution of block $I_k$ equals\n\\[\nC_k:=(b^k+b^{-k})\n      \\sum_{n\\in I_k}(-1)^{n}b^{-n}\n   =-\\frac{b}{b+1}\\,\\Bigl(b^{k-a_k}-b^{-3k-a_k}\\Bigr).\n\\]\n\nStep 6 –  Expressing the exponents compactly.  \nSince $a_k=k^{2}-k+1$,\n\\[\nk-a_k=-k^{2}+2k-1=-k(k-2)-1,\n\\qquad\n-\\,3k-a_k=-k^{2}-2k-1=-k(k+2)-1.\n\\]\nDefine\n\\[\nT_k:=b^{-k(k-2)-1}\\quad(k\\ge1).\n\\]\nThen\n\\[\nC_k=-\\frac{b}{b+1}\\bigl(T_k-T_{k+2}\\bigr).\n\\]\n\nStep 7 –  Telescoping.  \nThe sum of all $C_k$ is\n\\[\nS(b)=\\sum_{k=1}^{\\infty}C_k\n     =-\\frac{b}{b+1}\\sum_{k=1}^{\\infty}\\bigl(T_k-T_{k+2}\\bigr)\n     =-\\frac{b}{b+1}\\,\\bigl(T_1+T_2\\bigr),\n\\]\nbecause the series telescopes in steps of $2$.\n\nStep 8 –  The remaining two terms.  \n\\[\nT_1=b^{-1(1-2)-1}=b^{\\,0}=1,\n\\qquad\nT_2=b^{-2(2-2)-1}=b^{-1}.\n\\]\nTherefore\n\\[\nS(b)=-\\frac{b}{b+1}\\,(1+b^{-1})=-\\frac{b}{b+1}\\cdot\\frac{b+1}{b}=-1.\n\\]\n\nStep 9 –  Convergence.  \nBecause every term is bounded in modulus by\n\\((b^{\\langle n\\rangle}+b^{-\\langle n\\rangle})/b^{n}\\le\n2\\,b^{\\langle n\\rangle-n}\\),\nand $\\langle n\\rangle\\le\\sqrt n+\\tfrac12$, the general term is $O(b^{-\\tfrac12 n})$,\nensuring absolute convergence.  (Alternating signs are therefore not even\nneeded for convergence.)\n\nThus the series converges for every $b>1$ and its value is always $-1$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.774620",
        "was_fixed": false,
        "difficulty_analysis": "• The sign oscillation destroys absolute monotonicity and forces careful control of parity, making direct “copy–paste” of the original argument impossible.  \n• A new parameter $b$ is introduced; the solver must track all algebra with a symbol instead of a specific base and still recognise the telescoping pattern.  \n• A subtle parity argument (that $k^{2}-k+1$ is always odd) is essential; missing it breaks the telescoping completely.  \n• The geometric sum now has ratio $q=-1/b$, so one must manage complex signs and an extra denominator factor $1+1/b$, which does not appear in the original problem.  \n• Telescoping occurs only after the unexpected factorisation  \n\\((b^{k}+b^{-k})(1-b^{-2k})=b^{k}-b^{-3k}\\); finding this factorisation requires deeper algebraic insight than the straightforward cancellation in the original.  \n• Finally, the surprising independence of $b$ and the universal value $-1$ add a conceptual layer: the solver must explain why the entire one–parameter family collapses to the same constant.\n\nAll these features—oscillating signs, a free parameter, non-trivial parity considerations, and a subtler telescoping structure—make the enhanced variant substantially more intricate than either the original problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix a real parameter $b>1$.  \nFor every positive integer $n$ let $\\langle n\\rangle$ denote the integer that is closest to $\\sqrt n$ (if $\\sqrt n$ is half–integer, either neighbour may be chosen).  \nProve that the alternating series  \n\n\\[\nS(b)=\\sum_{n=1}^{\\infty}(-1)^{n}\\,\n      \\frac{\\,b^{\\langle n\\rangle}+b^{-\\langle n\\rangle}\\!}{b^{\\,n}}\n\\]\n\nconverges for every such $b$, and show that its sum is in fact  \n\\[\nS(b)=-1,\n\\qquad\\text{independent of }b.\n\\]",
      "solution": "Step 1 –  A convenient parametrisation of the set $\\{n\\ge 1:\\langle n\\rangle=k\\}$.  \nFor each positive integer $k$,\n\\[\n\\bigl(k-\\tfrac12\\bigr)^{\\!2}\\le n<\\bigl(k+\\tfrac12\\bigr)^{\\!2}\n\\Longleftrightarrow\nk^{2}-k+1\\;\\le\\; n\\;\\le\\; k^{2}+k .\n\\]\nHence\n\\[\nI_k:=\\{n:\\langle n\\rangle=k\\}\n      =\\{k^{2}-k+1,\\;k^{2}-k+2,\\ldots ,k^{2}+k\\},\n\\]\nso $|I_k|=2k$ and the smallest index in $I_k$ is\n\\(\na_k=k^{2}-k+1.\n\\)\n\nStep 2 –  Parity of the first index in $I_k$.  \nBecause $k^{2}-k$ is the product of consecutive integers, it is even; hence  \n\\(\na_k=k^{2}-k+1\n\\)\nis always odd.  Consequently\n\\(\n(-1)^{a_k}=-1\n\\)\nfor every $k$.\n\nStep 3 –  Writing $S(b)$ as a sum over the blocks $I_k$.  \nWithin block $I_k$ the factor $b^{\\langle n\\rangle}+b^{-\\langle n\\rangle}$\nis constant and equal to $b^k+b^{-k}$.  Thus\n\\[\nS(b)=\\sum_{k=1}^{\\infty}(b^k+b^{-k})\n      \\sum_{n\\in I_k}(-1)^{n}\\,b^{-n}.\n\\]\n\nStep 4 –  Evaluating the inner geometric sum.  \nPut $q:=-\\dfrac1b$ so that $|q|<1$ and\n\\(\n(-1)^n b^{-n}=q^{\\,n}.\n\\)\nFor $n=a_k+i\\;(0\\le i<2k)$ we have\n\\[\nq^{\\,n}=q^{\\,a_k+i}=q^{\\,a_k}\\,q^{\\,i}.\n\\]\nBecause $q^{\\,a_k}=(-1)^{a_k}b^{-a_k}=-\\,b^{-a_k}$ by Step 2,\n\\[\n\\sum_{n\\in I_k}(-1)^{n}b^{-n}\n     =-\\,b^{-a_k}\\sum_{i=0}^{2k-1}q^{\\,i}\n     =-\\,b^{-a_k}\\,\\frac{1-q^{2k}}{1-q},\n\\]\nand with $q=-\\tfrac1b$ this becomes\n\\[\n\\sum_{n\\in I_k}(-1)^{n}b^{-n}\n     =-\\frac{b^{-a_k}\\bigl(1-b^{-2k}\\bigr)}{1+1/b}\n     =-\\frac{b}{b+1}\\,b^{-a_k}\\bigl(1-b^{-2k}\\bigr).\n\\]\n\nStep 5 –  A crucial algebraic factorisation.  \nNote that\n\\[\n(b^k+b^{-k})\\bigl(1-b^{-2k}\\bigr)=b^k-b^{-3k}.\n\\]\nHence the total contribution of block $I_k$ equals\n\\[\nC_k:=(b^k+b^{-k})\n      \\sum_{n\\in I_k}(-1)^{n}b^{-n}\n   =-\\frac{b}{b+1}\\,\\Bigl(b^{k-a_k}-b^{-3k-a_k}\\Bigr).\n\\]\n\nStep 6 –  Expressing the exponents compactly.  \nSince $a_k=k^{2}-k+1$,\n\\[\nk-a_k=-k^{2}+2k-1=-k(k-2)-1,\n\\qquad\n-\\,3k-a_k=-k^{2}-2k-1=-k(k+2)-1.\n\\]\nDefine\n\\[\nT_k:=b^{-k(k-2)-1}\\quad(k\\ge1).\n\\]\nThen\n\\[\nC_k=-\\frac{b}{b+1}\\bigl(T_k-T_{k+2}\\bigr).\n\\]\n\nStep 7 –  Telescoping.  \nThe sum of all $C_k$ is\n\\[\nS(b)=\\sum_{k=1}^{\\infty}C_k\n     =-\\frac{b}{b+1}\\sum_{k=1}^{\\infty}\\bigl(T_k-T_{k+2}\\bigr)\n     =-\\frac{b}{b+1}\\,\\bigl(T_1+T_2\\bigr),\n\\]\nbecause the series telescopes in steps of $2$.\n\nStep 8 –  The remaining two terms.  \n\\[\nT_1=b^{-1(1-2)-1}=b^{\\,0}=1,\n\\qquad\nT_2=b^{-2(2-2)-1}=b^{-1}.\n\\]\nTherefore\n\\[\nS(b)=-\\frac{b}{b+1}\\,(1+b^{-1})=-\\frac{b}{b+1}\\cdot\\frac{b+1}{b}=-1.\n\\]\n\nStep 9 –  Convergence.  \nBecause every term is bounded in modulus by\n\\((b^{\\langle n\\rangle}+b^{-\\langle n\\rangle})/b^{n}\\le\n2\\,b^{\\langle n\\rangle-n}\\),\nand $\\langle n\\rangle\\le\\sqrt n+\\tfrac12$, the general term is $O(b^{-\\tfrac12 n})$,\nensuring absolute convergence.  (Alternating signs are therefore not even\nneeded for convergence.)\n\nThus the series converges for every $b>1$ and its value is always $-1$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.593078",
        "was_fixed": false,
        "difficulty_analysis": "• The sign oscillation destroys absolute monotonicity and forces careful control of parity, making direct “copy–paste” of the original argument impossible.  \n• A new parameter $b$ is introduced; the solver must track all algebra with a symbol instead of a specific base and still recognise the telescoping pattern.  \n• A subtle parity argument (that $k^{2}-k+1$ is always odd) is essential; missing it breaks the telescoping completely.  \n• The geometric sum now has ratio $q=-1/b$, so one must manage complex signs and an extra denominator factor $1+1/b$, which does not appear in the original problem.  \n• Telescoping occurs only after the unexpected factorisation  \n\\((b^{k}+b^{-k})(1-b^{-2k})=b^{k}-b^{-3k}\\); finding this factorisation requires deeper algebraic insight than the straightforward cancellation in the original.  \n• Finally, the surprising independence of $b$ and the universal value $-1$ add a conceptual layer: the solver must explain why the entire one–parameter family collapses to the same constant.\n\nAll these features—oscillating signs, a free parameter, non-trivial parity considerations, and a subtler telescoping structure—make the enhanced variant substantially more intricate than either the original problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}