absolutely convergent series and its properties

Question

Assume we have two absolutely convergent series $\{a_n\}^{\infty}_{n=1}$ and $\{b_n\}^{\infty}_{n=1}$ such that $\sum^{\infty}_{n=1}a_n=\sum^{\infty}_{n=1}b_n$ and $\lim_{n\to\infty}|{\frac{a_n}{b_n}}|=1$. Also $a_n$ and $b_n$ are identically different from zero for all $n$, satisfying also $a_n<a_{n-1}$ and $b_n<b_{n-1}$ for all $n$. What would be a sufficient condition for $a_n=b_n$ for all $n$?

Answer 1

evenConsider $\displaystyle a_n=\frac{1}{2^n}$ and $\displaystyle b_1=\frac7{16},b_2=\frac5{16},b_n=\frac{1}{2^n},n\ge 3.$ Then

$$\displaystyle \sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} \frac{1}{2^n}=2$$ and

$$\displaystyle \sum_{n=1}^{\infty} b_n=\frac7{16}+\frac5{16}+\sum_{n=3}^{\infty} \frac{1}{2^n}=\frac12+\frac14+\sum_{n=3}^{\infty} \frac{1}{2^n}=\sum_{n=1}^{\infty} \frac{1}{2^n}=2.$$

Both series satisfies the given conditions. This example can be generalized to get examples with $a_n\neq b_n$ for $n=1,\cdots,N,$ for any given $N\in\Bbb{N}.$ So, without very restrictive additional conditions your thesis won't hold.

Edit

Even assuming $a_n=b_n,\forall n\ge 3$ is not a sufficient condition. So, there is no any condition to guarantee $a_n=b_n,\forall n,$ except, of course, the trivial hypothesis $a_n=b_n,\forall n.$

Answer 2

(We know from the hypotheses that the sequences $(a_n)$ and $(b_n)$ are both decreasing and converge to 0.)

One sufficient condition would be that $a_n\le b_n$ for all n, since then $a_m\ne b_m$ for some m would imply that $\displaystyle\sum_{n=1}^{\infty}a_n < \sum_{n=1}^{\infty}b_n\;$, although there might be less restrictive conditions than this.