How to prove convergence of a series that has a recursively defined sequence?

Question

I've been banging my head against this one for a couple days ... relatively new to math and physics still, so I'm looking for some help here! This is actually coming out of a problem in a Quantum Mechanics textbook. I'm currently taking Analysis I to give you an idea of where I am in math.

In part of the problem, we are asked to show that the series shown below diverges as $|u|\to 1$, unless $\lambda = l(l+1)$ where $l=0,1,2...$

$$\sum_{k=0}^\infty a_k u^k$$

The sequence $a_k$ is defined by the recursion relation shown below

$$a_k\frac{(-k(k+1)+\lambda)}{(k+2)(k+1)}=a_{k+2}$$

I've tried comparison test, zero/divergence test, an argument that tries to use the contrapositive of the monotone bounded sequence convergence theorem, and an argument that tries to make use of the fact that the series we're given kind of looks like a geometric series... all to no avail.

Answer 1

If $\vert u \vert <1$ then the series converges the comparison test (the coefficients $a_k$ are bounded). On the other hand, if $\vert u \vert>1$ the series diverges by the divergence test (the coefficients decay slower than any exponential as we show below). Hence, we are left to consider what happens for $\vert u \vert=1$.

It seems to be false that for $\vert u \vert =1$ the series diverges (for $u\neq \pm i)$. It is unclear to me whether the series converges if $u=\pm i$.

Warm-up: u=1

Pick for simplicity $u=1$, then we are interested in the convergence of

$$ \sum_{k=1}^\infty a_k. $$

To show that $\sum_{k=1}^\infty a_k$ converges, it is enough to show that both

$$ \sum_{k=1}^\infty a_{2k}, \quad \sum_{k=1}^\infty a_{2k+1} $$ converge. I will focus on the first term, the second can be dealt with in a similar fashion.

We will show that $\sum_{k=1}^\infty a_{2k}$ satisfies the alternating series test and hence converges.

For this we note that

\begin{align*} \frac{-k(k+1)+\lambda}{(k+2)(k+1)} = \frac{-1-\frac{1}{k}+ \frac{\lambda}{k^2}}{1+\frac{3}{k}+\frac{2}{k^2}}. \end{align*} Now we expand the denominator and get \begin{align*} \frac{-1-\frac{1}{k}+ \frac{\lambda}{k^2}}{1+\frac{3}{k}+\frac{2}{k^2}} &= \left( -1-\frac{1}{k}+ \frac{\lambda}{k^2}\right) \left( 1-\frac{3}{k}+O(1/k^2)\right) \\ &= -1 + \frac{2}{k} + O(1/k^2). \end{align*}

Hence, we get (for $k$ large enough)

$$ \vert a_{k+2} \vert < \vert a_k \vert. $$

Furthermore, we can write for some $K$ large enough

$$ a_{2n+K} = a_K(-1)^n \underbrace{\prod_{j=0}^{n-1} \frac{(-\lambda+(K+2j)(K+2j+1)}{(K+2j+2)(K+2j+1)}}_{=: b_{2n+K}}.$$

We have $b_{2n+K}>0$. So the only thing that is left to check is that $\lim_{n\rightarrow \infty} b_{2n+K}=0$. It is a general fact that an infinite product of terms in $(0,1)$ converges to zero iff the series over the logarithms of the factors diverges (see How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$? and the links therein). In our case we have to deal with

$$ \prod_{j=0}^\infty \left( 1- \frac{2}{j} + O(1/j^2) \right).$$

We check that

$$ \sum_{j=0}^\infty \left(\frac{2}{j} + O(1/j^2) \right) = \infty $$ as $ \sum_{j=0}^\infty O(1/j^2) $ converges (by the comparison test with $C/j^2$) and the fact that the harmonic series diverges to $+\infty$. Thus, $$ \lim_{n\rightarrow \infty} b_{2n+K} = \prod_{j=0}^\infty \frac{(-\lambda+(K+2j)(K+2j+1)}{(K+2j+2)(K+2j+1)} = \prod_{j=0}^\infty \left(1 - \frac{2}{K+2j} + O(1/j^2)\right)= 0. $$

General $\vert u \vert =1$ where $u\neq \pm i$:

For a more general $u\in \mathbb{C}\setminus\{\pm i\}$ with $\vert u \vert=1$ we can resort to the Dirichlet test (see Explanation for the Proof of Dirichlet's test). In our case we have rewrite our series as

$$ \sum_{k=1}^\infty a_{2k+K} u^{2k+K} = \sum_{k=1} (-1)^k u^{2k+K} b_{2k+K}. $$ We have already shown that $(b_{2k+K})_{k\geq 1}$ is monotone decreasing and $\lim_{k\rightarrow \infty} b_{2k+K}=0$. Thus, we are left to check that there exists $M>0$ such that for all $N\in \mathbb{N}$ we have $$ \left\vert \sum_{k=1}^N (-1)^k u^{2k+K} \right\vert \leq M $$ We note that that the series in question is a geometric series and we have (for $u^2\neq -1,$ i.e. $u\neq \pm i$) $$ \sum_{k=1}^N (-1)^k u^{2k+K} = u^K \sum_{k=1}^N (-u^2)^k = -u^{K+2} \frac{1-(-u^2)^N}{1+u^2}. $$ Therefore, we have for $\vert u \vert =1$ and $u\neq \pm i$ $$ \left\vert \sum_{j=1}^N (-1)^k u^{2k+K} \right\vert \leq \frac{2}{\vert 1+u^2\vert}. $$ Hence, $$ \sum_{k=1}^\infty a_{2k} u^{2k} $$ converges by Dirichlet's criterion.

Special case $u=\pm i$:

I do not see on the spot whether the series converges in this case. I would assume here it starts to depend on the relation between $a_0$ and $a_1$. Maybe if we choose $a_0$ and $a_1$ in a suitable way we can use again Dirichlet's test with $\vert\sum_{j=1}^N (\pm i)^j \vert \leq \sqrt{2}.$

Answer 2

Write $\lambda = l(l+1)$ and $k = 2q + r$, $ r \in \{0, 1\}$. Then the recurrence relation takes the form

\begin{align*} a_{2q+r} &= - \frac{(2q+r-1+l)(2q+r-2-l)}{(2q+r)(2q+r-1)} a_{2(q-1)+r}. \end{align*}

Repeatedly applying this relation and then applying the Stirling's approximation, we get

\begin{align*} a_{2q+r} &= (-1)^q \frac{2^{2q} (q+\frac{r-1+l}{2})_{(q)} (q+\frac{r-2-l}{2})_{(q)} }{(2q+r)!} a_{r} \\ &= (-1)^q C_r \left( \frac{1}{q} + \mathcal{O}\left(\frac{1}{q^2}\right) \right) \quad \text{as } q \to \infty, \end{align*}

where $(x)_{(n)} = x(x-1) \cdots (x-n+1)$ is the falling factorial, $C_r$ is defined by

$$ C_r = \frac{\sqrt{\pi} \, a_{r}}{2^r (\frac{r-1+l}{2})! (\frac{r-2-l}{2})! }, $$

and we adopt the convention $\frac{1}{s!} = 0$ for $s \in \{-1, -2, -3, \ldots\}$. Consequently, we obtain the following result:

Conclusion. Let $|u| = 1$. For each $r \in \{0, 1\}$, we consider the series $$ \sum_{q=0}^{\infty} a_{2q+r} u^{2q+r}. $$

1. When $C_r = 0$, the series is eventually terminating, hence it converges absolutely. This condition occurs in each of the following scenarios:

   • $a_r = 0$, or
   • $l = p - r$ for some negative odd integer $p$, or
   • $l = p + r$ for some nonnegative even integer $p$.

   (I will leave the part of rephrasing this condition in terms of $\lambda = l(l+1)$ to others.)

2. When $C_r \neq 0$ and $u \notin \{ \pm i \}$, the series converges conditionally by the Dirichlet test.

3. When $C_r \neq 0$ and $u \in \{ \pm i \}$, the series diverges by the limit comparison with the harmonic series.

Answer 3

Throughout this posting, we assume that $\lambda\notin\{\ell(\ell+1):\ell\in\mathbb{Z}_+\}$. $b_k=a_{2k}$ $c_k=a_{2k-1}$ for $k\in\mathbb{N}$.

Notice that there is $K$ such that $b_{k+1}/b_k<0$ and $c_{k+1}/c_k<0$ for all $k\geq K$. Then, the terms of the series $\sum_{k\geq K}b_k$ and $\sum_{k\geq K}c_k$ alternate in sign. For $k\geq K$, \begin{align} -\frac{b_{k+1}}{b_k}=\frac{|b_{k+1}|}{|b_k|} &= \frac{4k^2+2k-\lambda}{4k^2+6k+2}=1-\frac{4k-2-\lambda}{4k^2+6k+2}=1-\frac{1}{k}+O\big(\tfrac1{k^2}\big)\\ -\frac{c_{k+1}}{c_k}=\frac{|c_{k+1}|}{|c_k|} &= \frac{4k^2-2k-\lambda}{4k^2+2k}=1-\frac{4k-\lambda}{4k^2+2k}=1-\frac{1}{k}+O\big(\tfrac1{k^2}\big) \end{align} It follows that $\big|\frac{b_{k+1}}{b_k}\big|\vee \big|\frac{c_{k+1}}{c_k}\big|<1$ for $k\geq K$ and so, the sequences $|b_k|$ and $|c_k|$ are eventually monotone decreasing and thus, they converge.

A simple calculation shows that $$\Big(k\big(1-\frac{|b_{k+1}|}{|b_k|}\big)-1\Big)\log k=\frac{(\lambda-2)k-2}{4k^2+6k+2}\log k\xrightarrow{k\rightarrow\infty}0<1 $$ and $$\Big(k\big(1-\frac{|c_{k+1}|}{|c_k|}\big)-1\Big)\log k=\frac{(\lambda-2)k}{4k^2+2k}\log k\xrightarrow{k\rightarrow\infty}0<1 $$ Thus, both $\sum_n|b_n|$ and $\sum_n|c_n|$ diverge by Bertrand's test. To consider the possibility of conditional convergence along the circle $|z|=1$, we determine the limits $\lim_k|b_k|$ and $\lim_n|c_k|$. Notice that $$\frac{|b_{k+K}|}{|b_K|}=\prod^{k+K-1}_{j=K}\Big(1-\frac1j+ O\big(\frac{1}{j^2}\big)\big) $$ Since $|b_k|$ converges and $\sum_j\Big(\frac{1}{j}+O\big(\frac{1}{j^2}\big)\Big)=\infty$, it follows that $\lim_kb_k=0$. The same argument shows that $\lim_kc_k=0$.

Putting things together, we have that the series $\sum_kz^{2k}b_k$ and $\sum_kz^{2k-1}c_k$ converge absolutely inside the unit disk $|z|<1$, and conditionally for all $z$ with $|z|=1$ (by Dirichlet's test) except for $z\in\{\pm i\}$ (this corresponds to the outcome of Bertrand's test).