Is it true that if $f(a_{n},s)\sim g(a_{n},s),\,n\rightarrow\infty$ then $\left|\sum_{n}f(a_{n},s)\right|\leq C\left|\sum_{n}g(a_{n},s)\right|,\,C>0?$

Question

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.

Answer 1

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 $1 \not< C\cdot0$.

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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.

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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 $1 \not< C\cdot0$.