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Is the following assertion true? Justify your answer. 􏰑􏰑

Suppose that we have series $\sum_{k=p}^∞ a_k$ and $\sum_{k=p}^∞ b_k$. Suppose also that $a_k = b_k$ for all but finitely many k. Then $\sum_{k=p}^∞ a_k$ converges if and only if $\sum_{k=p}^∞ b_k$ converges.

I'm really struggling to either prove/disprove this statement. Could someone please get me on the right track of starting? Is it related to subsequences or am I looking into the wrong section of this topic?

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    $\begingroup$ If it's a finite amount, you'll have a greatest $K$, so that the sequences are equal for all $k>K$. You can then split the partial sums $S_n$ in two for $n>K$. $\endgroup$ – dafinguzman Mar 27 at 14:08
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Hint:

  1. For any $N\in\mathbb N$, the series $$\sum_{n=1}^\infty a_n$$ converges if and only if the series $$\sum_{n=N}^\infty a_n$$ converges.
  2. You can prove (1) by looking at the definition of convergence for sums (the one with partial sums)
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Let the sets $A = \left\lbrace a_k \right\rbrace$ and $B = \left\lbrace b_k \right\rbrace$ denote those elements which are not equal for the sequence $\left\lbrace a_k \right\rbrace$ and $\left\lbrace b_k \right\rbrace$.

Removing these elements (which are only finitely many), let the new sequences obtained by $\left\lbrace a_k' \right\rbrace$ and $\left\lbrace b_k' \right\rbrace$. Observe that these two sequences are exactly the same ones. Hence, $\sum\limits_{k} a_k' = \sum\limits_{k} b_k'$, i.e., they converge or diverge together.

Now, adding finitely many elements will not change the nature of convergence (but only the limit, if the series is convergent). Hence, the two series mentioned by you converge or diverge together.

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First show the following : if $a_k = b_k$ for all but finitely many $k$, then there exists $N \in \mathbb N$ such that $a_n =b_n$ for all $n > N$ (argue by contradiction).

Now, let $\sum_{k=p}^N a_k - \sum_{k=p}^N b_k = P$. Note that $P < \infty$, being the difference of two real numbers. Show that $\sum_{k=p}^n a_k - \sum_{k=p}^n b_k = P$ for all $n > N$. Conclude that one series converges if and only if the other does , from the definition of the partial sums converging.

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  • $\begingroup$ how would i argue the first point by contraction? could you start me off? $\endgroup$ – beth Mar 27 at 14:59
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    $\begingroup$ Ok. Suppose that "there exists $N \in \mathbb N$ such that $a_n = b_n$ for all $n > N$" is false. What does this mean? Well, for any $N \in \mathbb N$, there is some $m > N$ such that $a_m \neq b_m$, right? Now, start with some $N_1$ such that $a_{N_1} \neq b_{N_1}$. Use the statement to conclude there is $N_2 > N_1$ with $a_{N_2} \neq b_{N_2}$. Now there is $N_3 > N_2$, then $N_4 > N_3$ and so on, all points at which $a_{N_k}$ and $b_{N_k}$ are not equal. Can you see that this contradicts the fact that $a_k$ and $b_k$ differ at only finitely many places? $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 at 15:02
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    $\begingroup$ Get back if you have got it. $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 at 15:11
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    $\begingroup$ That should be very easy : split the sum $\sum_{k=p}^n a_k$ into two parts : $\sum_{k=p}^N a_k$ and $\sum_{k=N+1}^n a_k$. Do this with $b_k$ also, and now take differences, noting $a_k= b_k$ for $k > N.$ $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 at 15:34
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    $\begingroup$ See, if the difference between two sequences is $P$, then one converges if and only if the other one does. To prove this, assume that one has a limit. What can the limit of the other one be? The difference is $P$ after some time, so the difference of the limits should also be $P$,right? That is, assume that $\sum_{k=p}^\infty a_k = L$. Prove that $\sum_{k=p}^\infty b_k = L-P$ using the definition of the partial sums converging. (Also, going to sleep, so cannot respond until tomorrow morning, apologies for that). $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 at 17:11

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