# why converges absolutely needed for that prove *?*

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.

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 $$\frac \epsilon2$$.

• Yes here is the problem. I use cauchy criterion but without concerned about their index. Thanks @Hagen Von Eitzen Sir Commented Jul 27, 2019 at 11:34

Your proof is not correct and it could not be since you did not use the assumption that the series converges absolutely. So, what you proved contradicts the Riemann rearrangement theorem.

The error in your proof lies in assuming that if you take $$N$$ large enough so that$$m\geqslant n\geqslant N\implies\left\lvert\sum_{k=n}^nb_k\right\rvert<\frac\varepsilon2,$$then, if you take some $$a_k$$'s all of which are such that $$k\geqslant N$$, then the absolute values of their sum will still be smaller than $$\frac\varepsilon2$$.

• Carlos Santos Sir I actually want to see if it is possible to prove this theorem without using converges absolutely condition Commented Jul 27, 2019 at 11:20
• No, you cannot, because, as I wrote in my answer, if the statement would hold without the assumption of absolute convergence, then the Riemann rearrangement thorem would be false. Commented Jul 27, 2019 at 11:27
• Now I can related why it's necessary Thanks @Jose Carlos Santos Sir Commented Jul 27, 2019 at 11:29

You can't prove it without assuming absolute convergence, because the statement is false without that assumption. For example, the alternating harmonic series $$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10}+\cdots$$ converges conditionally, but the rearrangement $$1-\frac12+\frac13-\frac14+\frac15-\frac16-\frac18+\frac17-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16}+\frac19-\frac1{18}-\frac1{20}-\frac1{22}-\frac1{24}-\frac1{26}-\frac1{28}-\frac1{30}-\frac1{32}+\frac1{11}-\cdots$$ does not converge, as it fails the Cauchy criterion: each block of consecutive negative terms has a sum $$\le-\frac14$$.

• Can you put these block highlighted as it seem confused to find $\leq \frac{-1}{4}$ @bof Sir. Commented Jul 27, 2019 at 16:44
• @emonhossain I don't understand your comment. What are you confused about? What is it you want highlighted?
– bof
Commented Jul 29, 2019 at 3:18