Let $f(x) = \sum\limits_{n=0}^{\infty} a_n x^n$ be finite or infinite, where $x$ is a real number and $(a_n)_n$ is an infinite sequence of positive integers. For any integer $n$, let $ (a_{L,n})_{L}$ be an infinite sequence of integers which equals $a_n$ for any $L$ large enough and for any L, let $$ f_L(x) = \sum\limits_{n=0}^{\infty} a_{L,n} x^n. $$ Is it always true that $$ \lim_{L \rightarrow \infty} f_L(x) = f(x)? $$Is there a famous theorem that can be applied from which the identity follows?

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    $\begingroup$ We can make so that $f_L(x) = \infty$ for any $L$, and any $x\neq 0$, but $f(x)$ is convergent. For example, $f(x)=\frac{1}{1-x}$ and $f_L(x)$ has it's tail as $\sum_{n=k+1}^\infty n!x^n$. $\endgroup$ – Jakobian Jul 28 '18 at 19:48

I can interpret "$(a_{n,L})$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_{n,L}=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_{n,L}=a_n$.

In the first case, your statement is false. Let $a_n=1$ and $a_{L,n}=a_n=1$ for $L>n$ and $a_{L,n}=0$ else. Then $\sum a_n x^n=\frac1{1-x}=f(x)$ and $\sum a_{L,n} x^n=\sum_{n=L+1}^\infty x^n = \frac{x^{L+1}}{1-x}=f_L(x)$. For $-1<x<1$, $f_L(x)\to 0\neq f(x)$.

In the second case, your statement is true, as $a_{L,n}=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_{L}(x)=f(x)$. Then of course $\lim_{L\to\infty} f_L(x)=f(x)$.

Or do you have another different definition in mind?

  • $\begingroup$ Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_{n,L} = a_n$ for any $n< L$, but I don't know if this might make the statement true. $\endgroup$ – QuantumLogarithm Jul 28 '18 at 20:12
  • $\begingroup$ If $a_{L,n} = 1$ for $L > n$ and $a_{L,n} = 0$ for $n \geqslant L$, then $f_L(x) = \sum_{n=0}^{L-1} x^n = (1-x^L)/(1-x) \to 1/(1-x) = f(x)$, as well, since $\lim_{L \to \infty}x^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something. $\endgroup$ – RRL Jul 28 '18 at 21:06
  • $\begingroup$ Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above. $\endgroup$ – Kusma Jul 28 '18 at 21:53

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