Prove that radius of convergence of a new series is no less that the old one

Question

Let $\sum_{k=0}^\infty a_kz^k\in \mathbb{C}[[z]]$ be a power series with radius of convergence $r$. Let $\{y_n\}_{n\ge 0}$ be defined by:

$$ y_0=1, \qquad\text{ and } \qquad y_n=\frac{1}{n}\sum_{j=1}^n a_{j-1}y_{n-j},\quad\forall n\ge 1.$$ Prove that the radius of convergence of $\sum_{k=0}^\infty y_kz^k$ is at least $r$.

Ideas: I have tried several ways and I cannot solve it.

An example that radius of $y$ series is greater than $a$ series:

    let $a_{2n}=1,a_{2n+1}=-1$, then $y_0=y_1=1$ and $y_n=0$ for $n\ge 2$

Answer 1

$A(z) =\sum_{k=0}^\infty a_kz^k$, $Y(z) =\sum_{k=0}^\infty y_kz^k $.

$y_n =\frac{1}{n}\sum_{j=1}^n a_{j-1}y_{n-j} =\frac{1}{n}\sum_{j=0}^{n-1} a_{j}y_{n-j-1} $ so $y_{n+1} =\frac{1}{n+1}\sum_{j=1}^{n+1} a_{j-1}y_{n+1-j} =\frac{1}{n+1}\sum_{j=0}^{n} a_{j}y_{n-j} $.

Therefore

$A(z)Y(z) =\sum_{n=0}^{\infty} z^n\sum_{j=0}^n a_jz_{n-j} =\sum_{n=0}^{\infty} z^n(n+1)y_{n+1} $.

This looks like a derivative, so let's try that.

$Y'(z) =\sum_{k=1}^\infty ky_kz^{k-1} =\sum_{k=0}^\infty (k+1)y_{k+1}z^{k} $.

Therefore $AY = Y'$ so $A =\dfrac{Y'}{Y} =(\ln(Y))' $ or $\ln(Y) =\int A$ or $Y =e^{\int A} $.

The radius of convergence of $\int A$ is at least that of $A$ since the coefficients are smaller and we can use the root test.

If $U(z) = e^{V(z)}$, $U(z)$ converges for any $z$ for which $V(z)$ converges, since we can just evaluate $V(z)$ and then take $exp$ of it.

Therefore, the radius of convergence of $Y(z)$ is at least that of $e^{A(z)}$.

Answer 2

Hint: if $g(z) = \sum_{k=0}^\infty a_k z^k$ and $f(z) = \sum_{k=0}^\infty y_k z^k$, find a differential equation for $f(z)$.

Answer 3

As was pointed out several times, this coefficient recursion belongs to the differential equation $y'(x)=a(x)y(x)$ where $a(x)=\sum a_kx^k$ and $y(x)=\sum y_kx^k$.

If $r$ is the radius of convergence of the first series $a(x)$, then by Cauchy-Hadamard, that is, the root test for power series $\limsup_n\sqrt[n]{|a_n|}=r^{-1}$, we get that for any $0<\rho<r$ there is a constant $C=\sup_|a_n|ρ^n<\infty$ so that for all indices $n$ $$ |a_n|\le Cρ^{-n}. $$ The solution of the auxillary IVP $$u'(x)=u(x)\sum Cρ^{-n}x^n=\dfrac{Cy(x)}{1-x/ρ}~\text{ with }~u(0)=1$$ is $$ u(x)=\left(1-\frac xρ\right)^{-Cρ}=\sum_{n=0}^\infty\binom{Cρ+n-1}{n}\left(\frac xρ\right)^n $$

Claim: The series $y_n$ is bounded by the coefficients of $u(x)$, $$ |y_n|\le u_n=\binom{Cρ+n-1}{n}ρ^{-n} $$

For $n=0$ this is true as $y_0=1=u_0$. Now assume that the claim is true for $y_0,...,y_{n-1}$. Then for $y_n$ we get $$ |y_n|\le\frac1n\sum_{j=0}^{n-1}Cρ^{-n+j+1}\,\binom{Cρ+j-1}{j}ρ^{-j}=\frac{Cρ^{-n+1}}{n}\sum_{j=0}^{n-1}\binom{Cρ+j-1}{j} $$

Now apply the generalized Pascal's triangle recursion to solve the last sum as a telescoping sum $$ \binom{x+1}{k+1}-\binom{x}{k}=\frac{x(x-1)\dots(x-k+1)}{k!}\left(\frac{x+1}{k+1}-1\right)=\binom{x}{k+1} \\~\\\implies \sum_{j=0}^{n-1}\binom{Cρ+j-1}{j}=1+\sum_{j=1}^{n-1}\left[\binom{Cρ+j}{j}-\binom{Cρ+j-1}{j-1}\right]=\binom{Cρ+n-1}{n-1} $$ so that finally $$ |y_n|\le\frac{Cρ}{n}\binom{Cρ+n-1}{n-1}ρ^{-n}=u_n. $$ By induction this now holds for all integer $n$.

As $\frac{u_n}{u_{n+1}}=\frac{n+1}{Cρ+n}ρ$ converges to $ρ$, the radius of convergence of $u$ is $ρ$ and thus for $y$ is at least $ρ$. As this is true for any $ρ<r$, the radius of convergence of $y$ is also at least $r$.

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See also the answers to similar topics in

• On the radius of convergence of solutions of analytic ODE's
• Series expansion of ODE solution