Let $f(x)$ be the ordinary generating function for the sequence $ \{\ a_0 , a_1, a_2,... \}\ $. Find the ordinary generating sequence for the following sequence: $$b_n = a_n + c \ \ \ , n \in \mathbb{N}_0$$

My attempt

I'm guessing $c$ is just any constant. I'm having trouble understanding the question however.

I solved these types of questions such as:

Find an ordinary generating function for $a_n = n$

I did this by writing out $a_n$ as $1+2+3+4...$ and then writing out the power series $A(x) = x + 2x^2 + 3x^3...$.

Now I could write:

$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$

As I can see this from the power series.

From here I just had to algebraically manipulate both sides to find the generating function which I wanted to look like the power series $$\sum_{n=0}^\infty nx^n$$

However I'm having trouble relating this to my current question!

Thank you.

  • $\begingroup$ I would like some more insight on this problem. I'm mainly having problems understanding what $a_n$ is in this problem. I only know that it is $f(x)$ but am I supposed to answer in terms of $f(x)$?? $\endgroup$
    – Paze
    Apr 2, 2014 at 17:58
  • $\begingroup$ It is not right to write $a_n$ as $1 + 2 + 3 + 4\dots$. The $a_n$ that is given is a single particular fixed finite value, for any given $n$: the formula $a_n = n$ means that $a_0 = 0$ and $a_1 = 1$ and $a_2 = 2$ and $a_3 = 3$ and so on. The infinite sum $1 + 2 + 3 + \dots$ should never appear; you should write down your $A(x) = x + 2x^2 + 3x^3 + \dots$ (which is correct) directly. $\endgroup$ Apr 2, 2014 at 18:17
  • $\begingroup$ Oh and in this current problem, it's not even the case that $a_n = n$. And $a_n$ is not $f(x)$ either. $f(x)$ is one single object that holds (encodes) all the separate $a_n$, for every $n$: it means that $f(x) = a_0 + a_1x + a_2x^2 + \dots$. And yes, you're supposed to answer in terms of $f(x)$. $\endgroup$ Apr 2, 2014 at 18:19
  • $\begingroup$ Can you please tell me the relationship between $A(x)$ and $a_n$ or guide me to an explanation? Thank you. $\endgroup$
    – Paze
    Apr 2, 2014 at 18:27
  • $\begingroup$ See my answer to your previous question: if the relationship between $A(x)$ and $a_n$ is not clear from it, ask again. In short, the relationship is that $A(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots$, so $A(x)$ is a single object that encodes all the number $a_n$: by hanging the $n$th number $a_n$ onto the term $x^n$. $\endgroup$ Apr 2, 2014 at 18:30

1 Answer 1


The generating function for $b_n$ is by definition

$$g(x)=\sum_{n=0}^\infty b_n x^n=\sum_{n=0}^\infty (a_n+c) x^n=\sum_{n=0}^\infty a_n x^n+c\sum_{n=0}^\infty x^n$$

You know each of the two functions on RHS..

  • $\begingroup$ I certainly know the right-most function, but $$\sum_{n=0}^\infty a_n x^n$$ I am not so sure about? It seems familiar but I am just getting into this topic now $\endgroup$
    – Paze
    Apr 2, 2014 at 17:26
  • $\begingroup$ Would this be the same as $$\sum_{n=0}^\infty n x^n$$ ? $\endgroup$
    – Paze
    Apr 2, 2014 at 17:33
  • 2
    $\begingroup$ It's your original generating function (what you called $f$). $\endgroup$
    – Rus May
    Apr 2, 2014 at 17:59
  • 2
    $\begingroup$ So is $f(x) + \frac{c}{1-x}$ an acceptable answer? $\endgroup$
    – Paze
    Apr 2, 2014 at 18:01
  • 1
    $\begingroup$ Sounds good to me. $\endgroup$
    – Rus May
    Apr 2, 2014 at 18:10

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