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Attempt 2:

Let’s refer to a sequence $\{x_n\}_{n=1}^\infty$ as increasing after a while if the following is true:

$(\exists N \in \mathbb N)(\forall n \in\mathbb N)$$(n \leq N$ $\rightarrow$ $x_{n+1} \geq x_n)$.

Write a formal proof that if sequences $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ are both increasing after a while, then the sequence $\{x_n + y_n\}_{n=1}^\infty$ is also increasing after a while. (Make use of the definition of increasing after a while)

Proof: We must prove that $(\exists N \in N)(\forall n \in N)(n \geq N \rightarrow x_{n+1} + y_{n+1} \geq x_n + y_n)$. Since $x_n$ is increasing, it follows that there exists $N_1$ such that for all $n$ in the natural numbers, if $n \geq N$, then $x_{n+1} \geq x_n$. Similarly, since $y_n$ is increasing, it follows that there exists $N_2$ such that for all $n$ in the natural numbers, if $n \geq N$, then $y_{n+1} \geq y_n$. Choose $N_3 = \max(N_1, N_2)$. Let $n$ be a natural number. Suppose $n \geq N$. It follows that $x_{n+1} + y_{n+1} \geq x_n +y_n$. This completes the proof.

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  • $\begingroup$ You want that $x_{n+1} \geq x_n$ and $y_{n+1} \geq y_n$ for $n \geq N(x,y)$ holds. For which $N$ do you know, the first inequality holds, and for which $N$ do you know the second inequality holds (noting the $N$ in your definitions as $N(x)$ and $N(y)$ respectively. $\endgroup$ – ctst Sep 19 '16 at 22:08
  • $\begingroup$ your notation causes confusion. $\endgroup$ – KonKan Sep 19 '16 at 22:11
  • $\begingroup$ You know that for n > N x_n increases but might not for n < M. For m > M y_n increases but m < M might not so what number should we choose. The larger one, the smaller one, some third choice? $\endgroup$ – fleablood Sep 19 '16 at 22:13
  • $\begingroup$ "Do I need to put n,m≥N,M instead?" It is not at all clear what such a notation would mean. Also you aren't adding $x_m + y_n$ you are adding $x_k + y_k$ so you need one variable $k > K$ for some variable. What must K be. If K < N then we don't know if $x_k$ is increasing for $N> k > K$. If K < M then we don't know $x_k$ is increasing $M> k >K$. So how can we solve this? what must K be? $\endgroup$ – fleablood Sep 19 '16 at 22:19
  • $\begingroup$ How does it look now? $\endgroup$ – Remy Sep 19 '16 at 22:29
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So if N and M are the numbers in the two sequences, simply choose the larger of the two and it will work in the definition of the sum of the two numbers.

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Choose $N:=\max\{N',N''\}$, where $N'$ and $N''$are the respective indices for $x_n$ and $y_n$.

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As you have mentioned in your proof, a sufficient condition for $ x_{n+1} + y_{n+1} \geq x_n + y_n$, is that both $x_{n+1}\geq x_n$ and $y_{n+1} \geq y_n$.

Since, for $(x_n)_{n=1}^{\infty}$, there exists $N_1 \in \mathbb{N}$, and for $(y_n)_{n=1}^{\infty}$, there exists an $N_2 \in \mathbb{N}$, such that the two sequences are increasing, choose $N = $max$\{N_1,N_2\}$ for the third sequence.

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  • $\begingroup$ Ah, okay. I will use this for my 2nd attempt. $\endgroup$ – Remy Sep 19 '16 at 22:12
  • $\begingroup$ How does it look now? $\endgroup$ – Remy Sep 19 '16 at 22:21
  • $\begingroup$ Your proof was correct earlier as well, it just needed this definition of $N$. $\endgroup$ – panini Sep 19 '16 at 23:07
  • $\begingroup$ Ok, thanks for the help! $\endgroup$ – Remy Sep 19 '16 at 23:11

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