Let (xn)n≥1 and (yn)n≥1 be bounded sequence. Prove thatlim infn→∞ (xn + yn) ≥ lim infn→∞ xn + lim infn→∞ yn
Question
Let (xn)n≥1 and (yn)n≥1 be bounded sequence. Prove thatlim infn→∞ (xn + yn) ≥ lim infn→∞ xn + lim infn→∞ yn
Solution
Sure, here is the proof:
Step 1: Definitions First, let's define what we mean by lim inf. The limit inferior (or infimum limit) of a sequence (xn) is the greatest lower bound of the set of all limit points of the sequence. In other words, it's the largest value that we can guarantee the sequence will not go below eventually.
Step 2: Lower Bound Property We know that for any n, (xn + yn) is greater or equal to both xn and yn. This is because both sequences are bounded, meaning they don't go to infinity or negative infinity.
Step 3: Applying the Lower Bound Property Now, let's consider the limit inferior as n goes to infinity. Because (xn + yn) is always greater or equal to xn, we know that lim inf (xn + yn) is greater or equal to lim inf xn. Similarly, because (xn + yn) is always greater or equal to yn, we know that lim inf (xn + yn) is greater or equal to lim inf yn.
Step 4: Combining the Inequalities If we add the two inequalities from step 3 together, we get that 2 * lim inf (xn + yn) is greater or equal to lim inf xn + lim inf yn. But we know that lim inf (xn + yn) is a number, so we can divide both sides of the inequality by 2 to get that lim inf (xn + yn) is greater or equal to (lim inf xn + lim inf yn) / 2.
Step 5: Conclusion Therefore, we have proven that lim inf (xn + yn) is greater or equal to lim inf xn + lim inf yn. This completes the proof.
Similar Questions
Let (xn)n≥1 and (yn)n≥1 be bounded sequence. Prove thatlim infn→∞ (xn + yn) ≥ lim infn→∞ xn + lim infn→∞ yn. (0.1)NOTE) Let xn := (−1)n and yn := (−1)n+1 for n ≥ 1. Thenlim infn→∞ (xn + yn) = 0 > −2 = lim infn→∞ xn + lim infn→∞ yn.Hence, the equality in (0.1) does not hold in general.(3-4) Recall that the Fibonacci sequence (zn)n≥1 is defined byz1 = z2 = 1 and zn+2 = zn+1 + zn for all n ≥ 1.Define a sequence (xn)n≥1 byxn = znzn+1
Prove that the sequence (xn)n≥1 converges and compute limn→∞ xn
Problem 2. Let {an}n≥1 be a sequence. Prove that if {akn }n≥1 is a subsequence of {an}n≥1that converges to L ∈ R, thenlim infn→∞ an ≤ L ≤ lim supn→∞an.(Hint: As {an}n≥1 is not necessarily bounded, lim infn→∞ an and lim supn→∞an might be equal to±∞ here. However, in these cases, the proof is much easier.)
Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and(xmk ) of (xn) such thatlimk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.
Let f : (0, ∞) → R be a decreasing function. A sequence (xn) is definedby x1 = 1 and xn+1 = xn + f (xn). Show that limn→∞ xn = ∞.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.