# If a_n < b_n for all n, then lim a_n < lim b_n

If $a_n$ and $b_n$ are convergent sequences and $a_n ≤ b_n$ for all n, show that a ≤ b where a and b are the limits of $a_n$ and $b_n$ respectively.

since a_n and b_n are convergent, there exists an N1 such that $|a_n - a| < ε$ for all n > N1 and an N2 such that $|bn - b| < ε$ for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take $a < a_n + ε ≤ b_n + ε$, but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?

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Prove by contradiction that a > b is impossible.

jbunniii
Homework Helper
Gold Member
Hint: if $a > b$, then you can find an $\epsilon > 0$ such that $a - \epsilon > b + \epsilon$.

Dick
Homework Helper
If $a_n$ and $b_n$ are convergent sequences and $a_n ≤ b_n$ for all n, show that a ≤ b where a and b are the limits of $a_n$ and $b_n$ respectively.

since a_n and b_n are convergent, there exists an N1 such that $|a_n - a| < ε$ for all n > N1 and an N2 such that $|bn - b| < ε$ for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take $a < a_n + ε ≤ b_n + ε$, but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?

thanks for your replies. i am working on it but still having trouble finding a contradiction. i know that a > b and a_n ≤ b_n along with $|a_n - a| < ε$ and $|b_n - b| ≤ ε$ will lead me to a contradiction, but i have been manipulating these all together and still can't find the contradiction.

i've also tried setting ε = (a-b)/4 like Dick said, but i'm not sure of how to get a contradiction from that either. i get that a_n < (5a-b)/4 and b_n < (a + 3b)/4 which doesn't tell me much. can you explain a little more about how the contradiction arises in this case?

Assuming $a > b$, and choosing $\epsilon$ so that $a - \epsilon > b + \epsilon > b_n$, $a_n > a - \epsilon > b + \epsilon > b_n$, which contradicts $a_n \le b_n$.

Since you are requested to prove that, it is best not to make any assumptions as to the rigor. The sketch I gave above is still not completely formal, since one needs to prove that it is possible to choose $\epsilon$ with the required property (even though it might seem obvious). Dick suggested one particular way, but it is still up to you to prove that such an $\epsilon$ satisfies $a - \epsilon > b + \epsilon$.