How in general one would solve an infinite series recurrence relations? For instance, I am interested to solve the following:

\begin{equation} \sum_{n=0}^{\infty} (-1)^{n} F(n)\{1-(\alpha n +\beta)^2\}=0 \end{equation}

Where $\alpha$ and $\beta$ are constants. In my case, $(\alpha n +\beta)^2$ makes everything harder. I do want to know if there is a technique for this type of recurrence relations. Does exponential generating functions work?

  • $\begingroup$ How is this a recurrence relation? It doesn't determine the individual $F(n)$, it is just a single global constraint. $\endgroup$ – Robert Israel Jul 6 '15 at 16:56
  • $\begingroup$ @RobertIsrael Yes, that's true, it's just an infinite series. I have my equation written in different forms. The above one is one of them. In another form $F(m)=\sum_{n=0}^{\infty} F(n)(\alpha n+\beta) P_{m,n}$. where $P_{m,n}$ is a complicated function of $m$ and $n$. I was able to use orthogonality of $P$'s to generate the above infinite series. I am not sure which one would be easier to solve! $\endgroup$ – Ben Jul 6 '15 at 17:00

I'm not sure if this helps, but there are (uncountably) infinitely many such $F(n)$, for multiple reasons, one being that if $F(n)$ is a solution then $cF(n)$ is also a solution for any constant $c$. But that doesn't even begin to describe the rich variety you can have. For example, all you need for $F(n)$ is that your sequence $F(n)(1-(\alpha n + \beta)^2)$ tends to $0$ in the "right way" to get the series formula to converge to zero, and you're done. So one of the only things you can really say at first glance is that $F(n)$ tends to zero fast enough so that your series expression terms are able to, i.e. $F(n) = o(1/n^2)$. You can construct such $F(n)$ iteratively that satisfy your infinite series equation, making arbitrary choices of certain constants along the way, but I'm not sure whether there is an elementary closed form formula for $F(n)$ that works. Perhaps that is the thrust of your question, how to find an elementary closed form solution. If so, you should update your question to reflect that (since there are iterative procedures that will create uncountably many solutions, even up to scaling, although they probably won't provide much insight as to how to get elementary closed form solutions.)

Also, if you are interested in an iterative procedure to construct $F(n)$, here's the basic idea. You choose $F(n)$ in increasing order of $n$ by choosing $F(n)$ to be anything you want for all $n \leq N$, and then for $F(n)$ for $n > N$ you choose $F(n)$ so that the partial sum going up to the $n$th term in your series, call is $S_n$, satisfies $|S_n| \leq c |S_{n-1}|$ for some universal constant $c < 1$. One way to do this, which almost gives an elementary closed formula, is to set $F(n) = (-1)^n / (1 - (\alpha n + \beta)^2)$ for $n = 1$, unless $\alpha + \beta = 1$ in which case you set $F(1) = 0$ and $F(2)$ according to the specified formula, and then for all $n > 1$, assuming you were able to set $F(1)$ non-zero, you set $F(n) = (-1)^{n+1} \cdot (1/2)^{n-1} / (1 - (\alpha n + \beta)^2)$ so that in essence, the first term is $1$ and the other terms add up to $- \sum_{n = 1}^\infty 1/2^n = -1$, so your series is zero. If $\alpha + \beta = 1$ you basically pull the same trick starting at $n = 2$ instead, with $F(1) = 0$.

  • $\begingroup$ Somewhat more simply, start with any sequence such that $\sum_n n^2 |F(n)| < \infty$ and then adjust $F(0)$, $F(1)$, $F(2)$ so that $\sum_n (-1)^n F(n) = \sum_n (-1)^n n F(n) = \sum_n (-1)^n n^2 F(n) = 0$ (note that the matrix $\pmatrix{1 & 0 & 0\cr -1 & 1 & -1\cr 1 & 2 & 4\cr}$ is invertible, so this can always be done). $\endgroup$ – Robert Israel Jul 6 '15 at 19:21
  • $\begingroup$ Thanks so much user2566092 and @Robert Israel. It was really helpful. $\endgroup$ – Ben Jul 6 '15 at 21:50

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