0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.