Reciprocal sums of squares of gibonacci polynomials

TheoremA telescoping series of powers of gibonacci polynomial ratios#●ready to prove

Let \(x > 0\) be real, let \(\alpha = (x + \sqrt{x^2 + 4})/2\), and let \(g_n\) be either the Fibonacci polynomials \(f_n(x)\), given by \(f_0 = 0\) and \(f_1 = 1\), or the Lucas polynomials \(l_n(x)\), given by \(l_0 = 2\) and \(l_1 = x\), where both satisfy \(z_{n+2} = x z_{n+1} + z_n\). Let \(k\), \(r\), \(t\), \(\lambda\) be positive integers and \(p\) a natural number with \(t \le 2p\). Then the series

\[\sum_{n=1}^{\infty} \left[ \left( \frac{g_{(2pn+t-2p)k+r}}{g_{(2pn+t-2p)k}} \right)^{\lambda} - \left( \frac{g_{(2pn+t)k+r}}{g_{(2pn+t)k}} \right)^{\lambda} \right]\]

converges unconditionally, and its sum is \(\left( g_{tk+r} / g_{tk} \right)^{\lambda} - \alpha^{\lambda r}\).

The source takes \(x\) to be an integer; the statement is made for real \(x > 0\), which contains the positive integers.

public theorem_wanted hasSum_gibonacciPoly_pow_ratio_telescoping (g : ℕ → ℤ[X])
    (hg : g = generalizedFibonacciPolynomial 0 1 X 1 ∨ g = generalizedFibonacciPolynomial 2 X X 1)
    {x : ℝ} (hx : 0 < x) {k p r t m : ℕ} (hk : 0 < k) (hr : 0 < r) (ht : 0 < t) (hm : 0 < m)
    (htp : t ≤ 2 * p) :
    HasSum (fun n : ℕ =>
        (aeval x (g ((2 * p * (n + 1) + t - 2 * p) * k + r)) /
            aeval x (g ((2 * p * (n + 1) + t - 2 * p) * k))) ^ m -
          (aeval x (g ((2 * p * (n + 1) + t) * k + r)) /
            aeval x (g ((2 * p * (n + 1) + t) * k))) ^ m)
      ((aeval x (g (t * k + r)) / aeval x (g (t * k))) ^ m -
        ((x + Real.sqrt (x ^ 2 + 4)) / 2) ^ (m * r))
Sources
  • Thomas Koshy, Sums Involving a Family of Gibonacci Polynomial Squares: Generalizations, The Fibonacci Quarterly 62 (2024), Article koshy11112023, Lemma 2.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremSeries over a family of Fibonacci polynomial squares#●ready to prove

Let \(x > 0\) be real, let \(\alpha = (x + \sqrt{x^2 + 4})/2\), and let \(f_n = f_n(x)\) be the Fibonacci polynomials, given by \(f_0 = 0\), \(f_1 = 1\) and \(f_{n+2} = x f_{n+1} + f_n\), evaluated at \(x\). Let \(k\), \(r\), \(t\) be positive integers and \(p\) a natural number with \(t \le 2p\). Then the series

\[\sum_{n=1}^{\infty} \frac{(-1)^{tk} f_r f_{2pk}}{f_{(2pn+t-p)k}^2 - (-1)^{tk} f_{pk}^2}\]

converges unconditionally, and its sum is \(f_{tk+r} / f_{tk} - \alpha^r\).

This is the Fibonacci-polynomial case of Theorem 3.1 of the source. The source takes \(x\) to be an integer; the statement is made for real \(x > 0\), which contains the positive integers.

public theorem_wanted hasSum_fibPoly_sq_family {x : ℝ} (hx : 0 < x) {k p r t : ℕ} (hk : 0 < k)
    (hr : 0 < r) (ht : 0 < t) (htp : t ≤ 2 * p) :
    HasSum (fun n : ℕ =>
        (-1) ^ (t * k) * aeval x (generalizedFibonacciPolynomial 0 1 X 1 r) *
            aeval x (generalizedFibonacciPolynomial 0 1 X 1 (2 * p * k)) /
          (aeval x (generalizedFibonacciPolynomial 0 1 X 1 ((2 * p * (n + 1) + t - p) * k)) ^ 2 -
            (-1) ^ (t * k) * aeval x (generalizedFibonacciPolynomial 0 1 X 1 (p * k)) ^ 2))
      (aeval x (generalizedFibonacciPolynomial 0 1 X 1 (t * k + r)) /
          aeval x (generalizedFibonacciPolynomial 0 1 X 1 (t * k)) -
        ((x + Real.sqrt (x ^ 2 + 4)) / 2) ^ r)
Sources
  • Thomas Koshy, Sums Involving a Family of Gibonacci Polynomial Squares: Generalizations, The Fibonacci Quarterly 62 (2024), Article koshy11112023, Theorem 3.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremSeries over a family of Lucas polynomial squares#●ready to prove

Let \(x > 0\) be real, let \(\Delta = \sqrt{x^2 + 4}\) and \(\alpha = (x + \Delta)/2\), and let \(f_n\) and \(l_n\) be the Fibonacci and Lucas polynomials evaluated at \(x\), given by \(f_0 = 0\), \(f_1 = 1\), \(l_0 = 2\), \(l_1 = x\) and the recurrence \(z_{n+2} = x z_{n+1} + z_n\) for both. Let \(k\), \(r\), \(t\) be positive integers and \(p\) a natural number with \(t \le 2p\). Then the series

\[\sum_{n=1}^{\infty} \frac{-(-1)^{tk} \Delta^2 f_r f_{2pk}} {l_{(2pn+t-p)k}^2 + (-1)^{tk} \Delta^2 f_{pk}^2}\]

converges unconditionally, and its sum is \(l_{tk+r} / l_{tk} - \alpha^r\).

This is the Lucas-polynomial case of Theorem 3.1 of the source. The source takes \(x\) to be an integer; the statement is made for real \(x > 0\), which contains the positive integers.

public theorem_wanted hasSum_lucasPoly_sq_family {x : ℝ} (hx : 0 < x) {k p r t : ℕ}
    (hk : 0 < k) (hr : 0 < r) (ht : 0 < t) (htp : t ≤ 2 * p) :
    HasSum (fun n : ℕ =>
        (-1) ^ (t * k) * (-(x ^ 2 + 4)) * aeval x (generalizedFibonacciPolynomial 0 1 X 1 r) *
            aeval x (generalizedFibonacciPolynomial 0 1 X 1 (2 * p * k)) /
          (aeval x (generalizedFibonacciPolynomial 2 X X 1 ((2 * p * (n + 1) + t - p) * k)) ^ 2 -
            (-1) ^ (t * k) * (-(x ^ 2 + 4)) *
              aeval x (generalizedFibonacciPolynomial 0 1 X 1 (p * k)) ^ 2))
      (aeval x (generalizedFibonacciPolynomial 2 X X 1 (t * k + r)) /
          aeval x (generalizedFibonacciPolynomial 2 X X 1 (t * k)) -
        ((x + Real.sqrt (x ^ 2 + 4)) / 2) ^ r)
Sources
  • Thomas Koshy, Sums Involving a Family of Gibonacci Polynomial Squares: Generalizations, The Fibonacci Quarterly 62 (2024), Article koshy11112023, Theorem 3.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremReciprocal sums of shifted squares of Fibonacci and Lucas polynomials (Koshy–Gao, Theorem 2.1)#●ready to prove

Reciprocal sums of shifted squares of Fibonacci and Lucas polynomials (Koshy–Gao, Theorem 2.1). Let \(f_n(x)\) and \(l_n(x)\) be the Fibonacci and Lucas polynomials, given by \(z_{n+2} = x z_{n+1} + z_n\) with initial values \(f_0 = 0\), \(f_1 = 1\) and \(l_0 = 2\), \(l_1 = x\). Let \(x > 0\) be real, let \(k\), \(p\), \(r\), \(t\) be positive integers with \(t \le 2p\), and let \(\alpha = (x + \sqrt{x^2 + 4})/2\). Then

\[ \sum_{n=0}^{\infty} \frac{(-1)^{tk} f_r(x) f_{2pk}(x)} {f_{(2pn+p+t)k}(x)^2 - (-1)^{tk} f_{pk}(x)^2} = \frac{f_{tk+r}(x)}{f_{tk}(x)} - \alpha^r \]

and

\[ \sum_{n=0}^{\infty} \frac{(-1)^{tk+1} (x^2 + 4) f_r(x) f_{2pk}(x)} {l_{(2pn+p+t)k}(x)^2 + (-1)^{tk} (x^2 + 4) f_{pk}(x)^2} = \frac{l_{tk+r}(x)}{l_{tk}(x)} - \alpha^r. \]

In the source the sums run over \(n \ge 1\), with \((2pn + t - p)k\) in place of \((2pn + p + t)k\); here the summation index is shifted to start at \(n = 0\).

public theorem_wanted tsum_gibonacciPoly_sq_sub_family {x : ℝ} (hx : 0 < x) {k p r t : ℕ}
    (hk : 0 < k) (hp : 0 < p) (hr : 0 < r) (ht₀ : 0 < t) (ht : t ≤ 2 * p) :
    ∑' n : ℕ, (-1) ^ (t * k) * aeval x (𝔣 r) * aeval x (𝔣 (2 * p * k)) /
        (aeval x (𝔣 ((2 * p * n + p + t) * k)) ^ 2 - (-1) ^ (t * k) * aeval x (𝔣 (p * k)) ^ 2) =
      aeval x (𝔣 (t * k + r)) / aeval x (𝔣 (t * k)) - ((x + √(x ^ 2 + 4)) / 2) ^ r ∧
    ∑' n : ℕ, (-1) ^ (t * k + 1) * (x ^ 2 + 4) * aeval x (𝔣 r) * aeval x (𝔣 (2 * p * k)) /
        (aeval x (𝔩 ((2 * p * n + p + t) * k)) ^ 2 +
          (-1) ^ (t * k) * (x ^ 2 + 4) * aeval x (𝔣 (p * k)) ^ 2) =
      aeval x (𝔩 (t * k + r)) / aeval x (𝔩 (t * k)) - ((x + √(x ^ 2 + 4)) / 2) ^ r
Sources
  • Thomas Koshy and Zhenguang Gao, Sums Involving a Family of Jacobsthal Polynomial Squares, The Fibonacci Quarterly 62 (2024), Article koshy12202023, Theorem 2.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials

TheoremReciprocal sums of shifted squares of Jacobsthal and Jacobsthal–Lucas polynomials (Koshy–Gao, Theorem 3.1)#●ready to prove

Reciprocal sums of shifted squares of Jacobsthal and Jacobsthal–Lucas polynomials (Koshy–Gao, Theorem 3.1). Let \(J_n(x)\) and \(j_n(x)\) be the Jacobsthal and Jacobsthal–Lucas polynomials, given by \(z_{n+2} = z_{n+1} + x z_n\) with initial values \(J_0 = 0\), \(J_1 = 1\) and \(j_0 = 2\), \(j_1 = 1\). Let \(x > 0\) be real, let \(k\), \(p\), \(r\), \(t\) be positive integers with \(t \le 2p\), let \(w = (1 + \sqrt{4x + 1})/2\), and write \(e_n = (2pn + t)k\). Then

\[ \sum_{n=0}^{\infty} \frac{(-1)^{tk} x^{e_n} J_r(x) J_{2pk}(x)} {J_{(2pn+p+t)k}(x)^2 - (-1)^{tk} x^{e_n} J_{pk}(x)^2} = \frac{J_{tk+r}(x)}{J_{tk}(x)} - w^r \]

and

\[ \sum_{n=0}^{\infty} \frac{(-1)^{tk+1} (4x + 1) x^{e_n} J_r(x) J_{2pk}(x)} {j_{(2pn+p+t)k}(x)^2 + (-1)^{tk} (4x + 1) x^{e_n} J_{pk}(x)^2} = \frac{j_{tk+r}(x)}{j_{tk}(x)} - w^r. \]

In the source the sums run over \(n \ge 1\), with \((2pn + t - p)k\) in place of \((2pn + p + t)k\) and exponent \((2pn + t - 2p)k\) in place of \(e_n\); here the summation index is shifted to start at \(n = 0\).

public theorem_wanted tsum_jacobsthalPoly_sq_sub_family {x : ℝ} (hx : 0 < x) {k p r t : ℕ}
    (hk : 0 < k) (hp : 0 < p) (hr : 0 < r) (ht₀ : 0 < t) (ht : t ≤ 2 * p) :
    ∑' n : ℕ, (-1) ^ (t * k) * x ^ ((2 * p * n + t) * k) * aeval x (𝔍 r) *
          aeval x (𝔍 (2 * p * k)) /
        (aeval x (𝔍 ((2 * p * n + p + t) * k)) ^ 2 -
          (-1) ^ (t * k) * x ^ ((2 * p * n + t) * k) * aeval x (𝔍 (p * k)) ^ 2) =
      aeval x (𝔍 (t * k + r)) / aeval x (𝔍 (t * k)) - ((1 + √(4 * x + 1)) / 2) ^ r ∧
    ∑' n : ℕ, (-1) ^ (t * k + 1) * (4 * x + 1) * x ^ ((2 * p * n + t) * k) * aeval x (𝔍 r) *
          aeval x (𝔍 (2 * p * k)) /
        (aeval x (𝔧 ((2 * p * n + p + t) * k)) ^ 2 +
          (-1) ^ (t * k) * (4 * x + 1) * x ^ ((2 * p * n + t) * k) * aeval x (𝔍 (p * k)) ^ 2) =
      aeval x (𝔧 (t * k + r)) / aeval x (𝔧 (t * k)) - ((1 + √(4 * x + 1)) / 2) ^ r
Sources
  • Thomas Koshy and Zhenguang Gao, Sums Involving a Family of Jacobsthal Polynomial Squares, The Fibonacci Quarterly 62 (2024), Article koshy12202023, Theorem 3.1.
Uses 1
Statement usesGeneralized Fibonacci polynomials
5 ready to prove

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