Prove that radius of convergence of a new series is no less that the old one 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 $n\ge 2$
 A: 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 $f(z)$.
A: $A(z)
=\sum_{k=0}^\infty a_kz^k$,
$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
$e^{A(z)}$.
A: 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 $r$.

See also the answers to similar topics in


*

*On the radius of convergence of solutions of analytic ODE's

*Series expansion of ODE solution
