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Let $b_n=a_n-a_{n+1}$.

We first assume that ${a_n}$ converges so ,$\lim(b_n)=\lim(a_n)-\lim(a_{n+1})$ , hence $\lim(b_n)=0$.

Now let $s_m$ and $s_n$ be the consecutive partial sums of ${b_n}$.

So $|s_n-s_m| = |a_{n+1}-a_{m}|=|a_{n+1}-L+L-a_m|< \epsilon$ for $N\ge M \ge M(\epsilon)$.So the series is cauchy convergent .

Now let us assume that the series is convergent so the sequence formed by the partial sums are convergent hence $s_M=a_1-a_{M+1}$ hence $-\lim(s_M) +a_1 = \lim(a_{M+1})$.So $A=-S+a_1$.

Can someone go through my attempt and point out my mistake instead of suggesting another answer.

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  • $\begingroup$ The Cauchy part is not clear, and I guess it's not even needed here. $\endgroup$ – Berci Apr 27 '20 at 16:16
  • $\begingroup$ Dont you need to show that series $b_n$ is convergent? $\endgroup$ – Antimony Apr 27 '20 at 16:20
  • $\begingroup$ Is it acceptable if I went through your proof line by line with errrors/ things done right? $\endgroup$ – Teresa Lisbon Apr 27 '20 at 16:42
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Let $b_n=a_n-a_{n+1}$.

We first assume that ${a_n}$ converges so ,$\lim(b_n)=\lim(a_n)-\lim(a_{n+1})$ , hence $\lim(b_n)=0$.

You've shown that if $b_i$ converges, then it converges to $0$, but you haven't shown that $\lim_{i \to \infty} b_i$ exists. If you're looking to show $b_n \to 0$, then similarly to what you've done for $s_m$ below, you want to argue $$\lvert b_n \rvert \leq \lvert a_n - a \rvert + \lvert a_{n+1} - a \rvert \leq 2 \varepsilon$$ for all $n > N$ for some $N$, via convergence of $a_n$.

However, you're trying to show that $\sum b_n$ converges, and knowing $b_n \to 0$ doesn't help you here.

Now let $s_m$ and $s_n$ be the consecutive partial sums of ${a_n}$.

So $|s_n-s_m| = |a_{n+1}-a_{m}|=|a_{n+1}-L+L-a_m|< \epsilon$ for $N\ge > M \ge M(\epsilon)$.So the series is cauchy convergent .

I'm not entirely clear on what you've assumed here, nor exactly what your definition of $s_n$ is. Is it $\sum a_n$ or $\sum (a_n - a_{n+1})$? I think you want the latter, but "consecutive partial sums" means $S_n$ and $S_{n+1}$ for some partial sums $S$, not "partial sum of differences of consecutive terms".

Now let us assume that the series is convergent so the sequence formed by the partial sums are convergent hence $s_M=a_1-a_{M+1}$ hence $-\lim(s_M) +a_1 = \lim(a_{M+1})$.So $A=-S+a_1$.

I believe this is your attempt at "$\sum (a_i - a_{i+1})$ converges $\implies a_n$ converges". This is mostly fine, but you should be careful when you write $\lim(a_{M+1})$ - you're trying to show that this exists, and here you're assuming it exists. I would be somewhat more careful and write this as $$a_{M+1} = a_1 - s_M \implies a_M \to a_1 - S$$

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It's a telescoping series.

$\sum_{k=1}^n (a_k - a_{k+1}) = a_1 - a_{n+1}$.

so $\lim\limits_{n\to \infty} \sum_{k=1}^n (a_k - a_{k-1}) = \lim\limits_{n\to \infty} (a_1 - a_{n+1}) = a_1 - \lim\limits_{n\to \infty} a_n$

Which converges and exists if and only if the $a_n$ converges.

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  • $\begingroup$ "Now let sm and sn be the consecutive partial sums of an." What does that mean why are there two of them? Then you do weird eqation where you have variables, $m,n, L, n+1, \epsilon, M,N,M(\epsilon)$ with NONE of them being introduced or defined. It looks like you are trying to do definition proof of a cauchy sequence buy just mumbling the phrases without actually specifying what they are. Seems like you are doing preposterously to much work..... $(a_1-a_2) + ..... = a_1 - a_{n+1}$ and that converges if and only if $a_n$ converges (which should be a proposition you have already proven). $\endgroup$ – fleablood Apr 27 '20 at 16:49

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