Sorry for the dumb question. Let $s=a+ib$ a complex number and $a_{n}\in\mathbb{C}$ a complex sequence. Assume that $$f(a_{n},s)\sim g(a_{n},s),\,n\rightarrow\infty,$$ and assume also $\sum_{n}f(a_{n},s),\,\sum_{n}g(a_{n},s)$ are convergent and $$f(a_{n},s),g(a_{n},s)\not\equiv0$$. I would like to prove that exist some $C>0$ such that $$\left|\sum_{n}f(a_{n},s)\right|\leq C\left|\sum_{n}g(a_{n},s)\right|.$$ My thought was that $$f(a_{n},s)\sim g(a_{n},s)\Rightarrow\sum_{n}f(a_{n},s)\sim\sum_{n}g(a_{n},s)$$ and so, using the definition of limit, $\forall \epsilon>0$ exists $N_{0}$ such that if $N>N_{0}$ then $$\left|\sum_{n=1}^{N}f(a_{n},s)\right|\leq\left(1+\epsilon\right)\left|\sum_{n=1}^{N}g(a_{n},s)\right|$$ so $$\left|\sum_{n=1}^{\infty}f(a_{n},s)\right|\leq\left(1+\epsilon\right)\left|\sum_{n=1}^{\infty}g(a_{n},s)\right|$$ but I'm not sure that works. Am I wrong? Thank you.

  • $\begingroup$ As I understand your question, I think the claim is incorrect. Take $f$ and $g$ as the sequences $2^{-n}$ and $0$. $\endgroup$ – The very fluffy Panda Mar 3 '17 at 18:28
  • $\begingroup$ @TheveryfluffyPanda You're right, I forgot to add the condition that there are not identically zero. $\endgroup$ – user422009 Mar 3 '17 at 18:34
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    $\begingroup$ Convergent sequences are bounded. So you can always find such a $C$. Where does this question come from? $\endgroup$ – The very fluffy Panda Mar 3 '17 at 18:52
  • $\begingroup$ @TheveryfluffyPanda They are functions of $s$. It came from a problem I'm studying. $\endgroup$ – user422009 Mar 3 '17 at 18:56

Suppose $f(a_1,s) = 1$, $f(a_n,s) = 0$ for $n > 1$ and $g(a_1,s) = 0$ for $n \ge 1$.

Then $\sum_n f(a_n, s) = 1$, but $\sum_n g(a_n,s) = 0$ and $1 \not< C\cdot0$.

The convergence of sequences cannot guarantee you anything about their sums, since it only takes into account the tail of the sequences and anything can happen in the first finitely many terms.

Edit: In response to your edit, the problem is still false, even with nonzero sequence values.

Let $f(a_n,s) = 2^{-n}$ and let $g(a_n,s) = 2^{-\left\lfloor \frac{n}{2} \right\rfloor - 1}(-1)^n $, i.e.

$$g(a_n,s) = \left\{\frac{1}{2},-\frac{1}{2},\frac{1}{4},-\frac{1}{4},\frac{1}{8},-\frac{1}{8},\frac{1}{16},-\frac{1}{16},\frac{1}{32},-\frac{1}{32},\ldots\right\}$$

Then $\left|\sum_{n=0}^\infty f(a_n, s)\right| = 2$, but $\left|\sum_{n=0}^\infty g(a_n,s)\right| = 0$ and $1 \not< C\cdot0$.

  • $\begingroup$ Thank you, I forgot to add that there are not identically zero. $\endgroup$ – user422009 Mar 3 '17 at 18:33
  • $\begingroup$ Sorry the dumb question, but I can bound a finite sum of terms with a constant. Am I wrong? $\endgroup$ – user422009 Mar 3 '17 at 18:45
  • $\begingroup$ @user422009 See my edit. $\endgroup$ – Alexis Olson Mar 3 '17 at 18:52
  • $\begingroup$ I didn't say that the series converges absolutely, I say they are only convergent. And $F,G$ are functions of $s$. $\endgroup$ – user422009 Mar 3 '17 at 18:55
  • $\begingroup$ @user422009 If they aren't absolutely convergent, then it doesn't make sense to compare their absolute sums, unless you mean partial sums. $\endgroup$ – Alexis Olson Mar 3 '17 at 18:56

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