Theorem: If $\sum_{k=1}^{\infty} a_k$ converges absolutely, then any rearrangement of this series converges to the same limit.
I will try to prove without using converges absolutely condition.
proof: Assume $\sum_{k=1}^{\infty} a_k$ converges absolutely to A. Let $\sum_{k=1}^{\infty} b_k$ be a rearrangement of $\sum_{k=1}^{\infty} a_k$ and $$t_m=\sum_{k=1}^{m} b_k=b_1+b_2+\cdots+b_m$$
for the partial sums of the rearranged series. Thus we want to show that $(t_m)\rightarrow A$. Let $\epsilon >0$ and choose $N_1$ such that $|s_n-A|<\frac{\epsilon}{2}$ $(\text{By hypothesis})$ for all $n\geq N_1$. Again using cauchy criterion for series there exists an $N_2$ $\in \mathbb{N}$ such that whenever $n>m\geq N_2$ $$\left|\sum_{m+1}^{n} a_k\right|<\frac{\epsilon}{2}$$ Now take $N=\max\{ N_1,N_2\}$. we know that the finite set of terms $\{a_1,a_2,a_3,\cdots,a_N \}$ must all appear in the rearranged series, and we want to move far enough out in the series $\sum_{n=1}^{\infty} b_n$ so that we have included all of these terms. Thus choose M=max{$f(k):1\leq k\leq N$ }. It should now be evident that if $m\geq M$, then $(t_m-s_N)$ consists of a finite set of terms. And so
\begin{align*}
|t_m-A|=|t_m-s_N+s_N-A|\leq |t_m-s_N|+|s_N-A|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{align*}
whenever $m\geq M$.
Is my proof is correct $?$ If not then where is the problem and why converges absolutely needed at all. Any explanation or solution will be appreciated.
Thanks in advance .
Best Answer
The "finite set of terms" making up $(t_m-s_N)$ need not be consecutive, i..e, their sum need not be of the form $\sum_{k_1}^{k_2}a_k$ and thus need not be bounded by $\frac \epsilon2$.