The series $\displaystyle\sum^\infty_{n=1}a_n$ converges absolutely.

Prove that for every polynomial $p(x)=p_0+p_1x+...+p_dx^d$, $\ \displaystyle\sum^\infty_{n=1}p(a_n)$ converges iff $p_0=0$.

What I did:


Let's take a look at $\displaystyle\sum^\infty_{n=1}\sum^d_{i=1}p_i(a_n)^i$, we know $a_n$ converges, raising it to an exponent won't change it's convergence and scaler multiplication won't either, so all 'elements' in the sum except for $p_0$ converge.

$\displaystyle\sum^\infty_{n=1}p_0$ will always diverge unless $p_0 = 0$.

  • $\begingroup$ You've pretty much proved it yourself. I don't understand what more you're asking for. I could write out a proof myself but it'd essentially be what you've got here but more verbose. $\endgroup$ Apr 29, 2014 at 10:50
  • $\begingroup$ @Stijn just wanted to verify I wasn't wrong. $\endgroup$
    – GinKin
    Apr 29, 2014 at 10:51
  • 1
    $\begingroup$ Not as far as I can see. The only thing I'd really add is making more of a fuss about the absolute convergence. If the series over $a_n$ is conditionally convergent then it's conceivable that the sum over, say $a_n ^2$ diverges. $\endgroup$ Apr 29, 2014 at 10:53
  • $\begingroup$ @Stijn: I believe you can make your comments into an answer. $\endgroup$
    – RghtHndSd
    Apr 29, 2014 at 12:12

1 Answer 1


You're essentially right. The only thing to be careful of is dealing with $\sum_{n=1} ^\infty \sum_{k=1} ^d p_k a_n ^j$. The point is \begin{align} \sum_{n=1} ^\infty \sum_{k=1} ^d p_k a_n ^j &= \sum_{k=1} ^d p_k \sum_{n=1} ^\infty a_n ^j \\ &\leq \sum_{k=1} ^d p_k \sum_{n=1} ^\infty |a_n|^k \\ &\ll \sum_{k=1} ^d p_kS_k \\ &\leq d \max_{1 \leq k \leq d} p_k S_k < \infty \end{align} where $S_k$ is finite for all $k$. Using the absolute converge is crucial here as, say, $\sum_{n=1} ^\infty (-1)^n n^{-1/2}$ converges but, obviously, $\sum_{n=1} ^\infty n^{-1}$ diverges.

This fact puts all the burden of convergence on the sum over the $p_0$ and, as you noted, that is infinite if and only if $p_0 = 0$.

  • $\begingroup$ What is $S_k$ ? $\endgroup$
    – GinKin
    Apr 29, 2014 at 15:48
  • $\begingroup$ $S_k$ is just the infinite sum over $|a_n|$ but in a neater form. We know that it's finite because of absolute convergence. $\endgroup$ Apr 29, 2014 at 16:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .