Prove that a sequence defined by a recurrence relation converges

Question

Consider the following recurrence relation:

$$ a_i = \frac{i+2}{2} \cdot \left(\frac{i}{i+1} - \sum_{j=1}^{i-1} \frac{2 a_j}{2i - j + 2}\right). $$

The first ten terms are:

• $0.75$
• $0.733333333333$
• $0.728174603175$
• $0.725850340136$
• $0.72458353375$
• $0.723808024231$
• $0.723294499472$
• $0.722934619774$
• $0.722671361045$
• $0.722472200516$.

It is natural to conjecture that the sequence converges. How can one prove it? Actually, for my application it would be enough to prove convergence in the sense of Cesàro.

Some partial results.

• It is easy to see that if $a_i$ converges, then the limit is equal to $1 / \ln 4 \approx 0.722$.
• If we assume that for a very large $i$ we have $a_j = 1 / \ln 4$ for all $j < i$, then we get $$ a_i = \frac{7}{8 \ln 2} - \frac{1}{2} + o(1), $$ which is strictly larger than $1 / \ln 4$. Thus, we must use the fact that $a_i$ does not converge too fast.
• One can check empirically, that $a_i - 1 / \ln 4$ behaves roughly like $i^{-\alpha}$, where $\alpha$ is around $1.6$. Probably, with some effort one can even find the exact value of $\alpha$, but it would be great to do things more systematically.

Answer 1

(Not an answer.) A wider setting is to consider sequences $(a_n)_{n\geqslant1}=(a_n^{c,\mathbf b})_{n\geqslant1}$ defined recursively by the identities $$ \sum_{k=1}^n\frac{a_k}{n+1-ck}=b_n, $$ for every $n\geqslant1$, for some given parameter $c$ in $(0,1)$ and some sequence $\mathbf b=(b_n)_{n\geqslant1}$ converging to some limit $\beta$. Thus, for example, $$ a_1=(2-c)b_1,\quad a_2=(3-2c)\left(b_2-\frac{2-c}{3-c}b_1\right),\ \cdots $$ Basic algebraic manipulations show that the sequence in the post is the case when $c=1/2$ and $b_n=n/(n+1)$ for every $n$, thus $\beta=1$.

Conjecture A For every $(c,\mathbf b)$ as above, the sequence $(a_n^{c,\mathbf b})_{n\geqslant1}$ converges.

When $(a_n)_{n\geqslant1}$ converges, then, as mentioned in the post when $c=1/2$, its limit is $\beta/\vartheta(c)$, where $$ \vartheta(c)=-\frac{\log(1-c)}c=\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{1}{n+1-ck}. $$ In the question, $\vartheta(c)=2\ln2$ and $\beta=1$, and indeed $\beta/\vartheta(c)=1/(2\ln2)$.

A first line of attack of conjecture A would be to show that the convergence of $(a_n)_{n\geqslant1}$ is asymptotic with respect to $\mathbf b$ in the sense that changing the values of $(b_n)_{n\leqslant N}$ does not change the convergence of $(a_n)_{n\geqslant1}$, for every finite $N$. Then one could probably assume without loss of generality that $b_n=1$ for every $n$ by linearity of the transformation $\mathbf b\mapsto(a_n^{c,\mathbf b})_{n\geqslant1}$ (but the (lack of) monotonicity might be a problem here...), and solve this case.

A remark is that one is in fact studying the inverse of the infinite lower triangular invertible matrix $(M^c_{n,k})_{n\geqslant1,k\geqslant1}$ defined by $M^c_{n,k}=0$ for every $1\leqslant n\leqslant k-1$ and, for every $1\leqslant k\leqslant n$, $$ M^c_{n,k}=\frac1{n+1-ck}. $$ Considering the infinite lower triangular matrix $(T^c_{n,k})_{n\geqslant1,k\geqslant1}$ which is the inverse of $(M^c_{n,k})_{n\geqslant1,k\geqslant1}$, one might be led to prove that $T^c_{n,k}\to0$ for every fixed $k$ when $n\to\infty$, and that $$ \lim\limits_{n\to\infty}\sum_{k=1}^nT^c_{n,k}=\frac1{\vartheta(c)}. $$ A still more general formulation including our problem (after a small modification which we omit) is as follows: choose some (say, positive continuous) function $u$ on $[0,1]$ and define recursively some sequence $(a_n)_{n\geqslant1}=(a_n^{u,\mathbf b})_{n\geqslant1}$ by the identities $$ \frac1n\sum_{k=1}^nu\left(\frac{k}n\right)a_k=b_n, $$ for every $n\geqslant1$. The case in the post is when $$ u(t)=\frac1{1-ct}. $$

Conjecture B For every $(u,\mathbf b)$ as above, if $\mathbf b$ converges, the sequence $(a_n^{u,\mathbf b})_{n\geqslant1}$ converges.

If $\mathbf b$ converges to $\beta$ and if $(a_n^{u,\mathbf b})_{n\geqslant1}$ converges, then $(a_n^{u,\mathbf b})_{n\geqslant1}$ converges to $\beta/\Theta(u)$, where $\displaystyle\Theta(u)=\int_0^1u(t)\,\mathrm dt.$