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Searching for some help finishing the following proof. If we let $(a_n)$ and $(b_n)$ be sequences, and suppose there exists $N\in N$ such that $a_n=b_n$ for all $n>N$. I need to prove that $\sum a_n$ converges if and only if $\sum b_n$ converges.

My workings so far:

So I know I need to show that if $\sum a_n$ converges if and only if $\sum b_n$ converges. So I must show
1)$\sum a_n$ -> $\sum b_n$ and
2) $\sum b_n$ -> $\sum a_n$

So beginning with 1) $\sum a_n = \sum b_n + c$

If I can prove $S_n = S_n' + c$ then I can try and prove $\sum a_n = lim S_n = limS_N' + C $with
$S_n= a_1+a_2+a_3+...+a_N+a_{N+1}+...+a_n$ and
$S_n'= b_1+b_2+b_3+....+b_N+b_{N+1}+...+bn$

I know that $a_{N+1}=b_{N+1}$ and $a_n=b_n$ Which gives $S_n-S_n'= c+ 0 $ which proves $S_n = S_n' + c$ but how do I prove the second part of $\sum a_n = lim S_n = limS_N' + C $ which will then show that if $\sum a_n$ converges then $\sum b_n$ converges.

I am not sure if my logic here is correct or not, looking for some help with finishing this result and then proving that if $b_n$ converges then $a_n$ must also.

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Because $S_n=S_n'+c$ and the convergence of $(S_n)$ is assumed, then $S_n'=S_n-c$ converges. So, $\sum b_n$ is a convergent series. Since $S_n\to\sum a_n$, then $\sum b_n=\sum a_n-c$.

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  • $\begingroup$ is this added on to my proof? $\endgroup$ – user123 Jan 19 '17 at 20:59
  • $\begingroup$ Yes, of course. I have the impression that the final step was a trouble to you. $\endgroup$ – szw1710 Jan 19 '17 at 21:00
  • $\begingroup$ so your answer shows that$\sum a_n = lim S_n = limS_N' + C $ which will then show that if $\sum a_n$ converges then $\sum b_n$ converges? $\endgroup$ – user123 Jan 19 '17 at 21:03
  • $\begingroup$ and thanks again $\endgroup$ – user123 Jan 19 '17 at 21:03
  • $\begingroup$ Yes, this shows it. $\endgroup$ – szw1710 Jan 19 '17 at 21:06

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