suppose we have the following two sequences

$$\alpha_k = (k-1)\left(1-\frac {1}{1+(k+1)l}\right) \quad , k \geq 2$$ $$\beta_k = (k-1)\left(1+\frac {1}{1+(k-1)l}\right) \quad , k \geq 2$$

where $l$ is a positive constant and define the sequence $c_k$ recursively by: $$c_2 = - 1/\beta_2 $$ $$c_3 = 0 $$ $$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1} \quad , k \geq 3 $$

it is not hard to see that this would give $$c_2 = - 1/\beta_2$$ $$c_{2k} = -\frac{\alpha_2}{\beta_2 }\cdot\frac{\alpha_4}{\beta_4 }\cdot\cdot\cdot \frac{\alpha_{2k-2}}{\beta_{2k-2} }\cdot \frac{1} {\beta_{2k}} \quad , k \geq 2$$

$$c_{2k+1} = 0 \quad , k \geq 1 $$

apparently we must have $d_k = c_{2k} \sim k^{-(1+1/l)}$ but I have no idea how to show this. can anyone shed some light on this?

also a similar question for the recursion $$c_2 = - 1/\beta_2 $$ $$c_3 = \frac{\gamma_2}{\beta_3}c_2 $$ $$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1} + \frac{\gamma_k}{\beta_{k+1}}c_k \quad , k \geq 3 $$

where $$\gamma_k = \sigma \frac{l(k^3-k)}{1+kl} \quad , k \geq 2$$ $\sigma$ being also a positive constant

How would the asymptotics look like in this case?


2 Answers 2


For the first sequence, $$d_k=\frac{\alpha_{2k-2}}{\beta_{2k}}d_{k-1}=\frac{k-3/2}{k+1/2-1/\ell}d_{k-1}$$ so one has $$d_k=C\frac{\Gamma(k-1/2)}{\Gamma(k+3/2-1/\ell)}\ ,$$ the constant $C$ being determined by the initial condition $d_1$, namely $$C=d_1\frac{\Gamma(5/2- 1/\ell)}{\Gamma(1/2)}\ . $$ Recall that $\Gamma(x+a)=x^a\Gamma(x)(1+o(1))$ as $x\to+\infty$, so $$d_k=Ck^{-1-1/\ell}(1+o(1)). $$

For the second sequence, $$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }c_k+\frac{k-2}{k+2/\ell}c_{k-1}$$ which implies $c_{k-1}/c_k=O(1/k)$; if we plug this in the recursion again we have $$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }(1+O(1/k^2))\ , $$ whence $$c_k= A\frac{\Gamma(k+1)\Gamma(k-1)}{\Gamma(k+2/\ell) }(1+o(1))\ ,$$ because the infinite product of $1+O(1/k^2)$ is convergent. By the Stirling formula $$c_k= B k^{k-1/2+2/\ell}e^{-k} (1+o(1))$$ for a certain constant $B$.

  • $\begingroup$ is it just me or is the sign in the power of $k$ wrong in your answer for the first question? $\endgroup$ Jan 1, 2016 at 0:17
  • $\begingroup$ it is… fixed now $\endgroup$ Jan 1, 2016 at 7:37

The first factor telescopes completely, and the second becomes easier by combining $\alpha_{2\kappa}$ with $\beta_{2\kappa+2}$, so $$ c_{2k} = \frac{1+\ell}{(2k-1)(2+\ell)}\prod_{\kappa=1}^k \left(1-\frac{2}{2+(2\kappa+1)\ell}\right). $$ Now take logarithms, and apply Euler-MacLaurin summation.

  • $\begingroup$ Do you mean by applying the Eucler-Maclaurin summation that I should replace the sum of logarithms by an integral over a logarithm? $\endgroup$ Dec 31, 2015 at 16:48
  • $\begingroup$ If $f$ is a smooth function, then $\int f(t)$ is the first approximation to $\sum f(n)$. Euler-Maclaurin gives a sequence of correction terms, involving higher and higher derivatives of $f$. When applied to a sum of length $N$, using $k$-th derivatives usually gives an error of size $k!^C N^{-k}$, so taking $k$ too big does not pay off, but most of the time one or two terms suffice. $\endgroup$ Jan 1, 2016 at 11:09
  • $\begingroup$ Anyway, Majer's answer is better. The final result is the same, as the derivation of Stirling involves Euler-Maclaurin, but this way you save yourself a lot of work. $\endgroup$ Jan 1, 2016 at 11:13

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