Hello everybody, I'm not sure how to solve the following problem/proof:
The sequence (a_n) is defined by a_1 = 1/2 and
a_n = a_(n - 1)^2 + a_(n - 1) for n >= 2.
Prove that
1\(a_1 + 1) + 1\(a_2 + 1) + ... + 1\(a_n + 1) < 2 for all n >= 1.
I don't know how to start this one. I have always been weak with sequences and series :( If there's a particular formula that would help me with solving this, please share that much with me so I can attempt to figure it out on my own afterwards. If there isn't a formula that you can think of, that's okay too! Any help is appreciated.
See if you can show that a_(n+1) > 3/2 a_n
Having that, now you know that
1/(a(n+1)+1) < 1/a(n+1) < 2/3 1/a_n
That means that each term of the sum is less than the corresponding term in the geometric sequence
a = 2/3
r = 2/3
The infinite sum of that is
a/(1-r) = 2/3 * 1/(1 - 2/3) = 2/3 * 3 = 2
So since the infinite sum is 2, and all terms are positive, then each partial sum is less than 2.
Thank you. I'll get to work on that immiediately.
To prove the inequality 1/(a1 + 1) + 1/(a2 + 1) + ... + 1/(an + 1) < 2 for all n >= 1, we will use mathematical induction.
Step 1: Base Case
First, let's prove the statement for the base case n = 1.
We have a1 = 1/2, so the inequality becomes:
1/(1/2 + 1) < 2
Simplifying gives: 2/3 < 2, which is true.
Step 2: Induction Hypothesis
Assume that the statement is true for some arbitrary positive integer k, i.e.,
1/(a1 + 1) + 1/(a2 + 1) + ... + 1/(ak + 1) < 2.
Step 3: Inductive Step
Now, we need to prove that if the statement is true for k, then it is also true for k + 1.
Consider the sum S(k) = 1/(a1 + 1) + 1/(a2 + 1) + ... + 1/(ak + 1).
To prove the inequality for k + 1, we need to show that:
S(k + 1) = S(k) + 1/(ak + 2 + 1) < 2.
Using the induction hypothesis, we can rewrite S(k) as less than 2:
S(k) < 2.
Now, let's analyze the term 1/(ak + 2 + 1).
We can see that a(k+1) = ak^2 + ak = ak (k + 1), based on the given recursive formula.
Therefore, 1/(ak + 2 + 1) = 1/(ak^2 + ak + 1) = 1/(ak (k + 1) + 1).
Since ak > 0 and k + 1 > 1 for all positive integers k, it follows that ak (k + 1) + 1 > ak + 1.
Thus, we have 1/(ak + 2 + 1) < 1/(ak + 1).
Now, let's rewrite S(k + 1) as a sum of two terms:
S(k + 1) = S(k) + 1/(ak + 2 + 1).
We know that S(k) < 2, and from the above inequality, 1/(ak + 2 + 1) < 1/(ak + 1).
Therefore, we can conclude that:
S(k + 1) < 2 + 1/(ak + 1).
Now, we will simplify the expression 2 + 1/(ak + 1).
Multiplying the numerator and denominator of 1/(ak + 1) by (ak + 1), we get (ak + 1)/(ak + 1)^2 = 1/(ak + 1).
Thus, 2 + 1/(ak + 1) = 2 + (ak + 1)/(ak + 1) = (2ak + 3)/(ak + 1).
To prove S(k + 1) < (2ak + 3)/(ak + 1) for all k, we can rearrange the inequality:
S(k + 1) < (2ak + 3)/(ak + 1)
S(k + 1) < 2 - 1/(ak + 1) + 3/(ak + 1)
Now, notice that S(k + 1) < S(k) + 1/(ak + 1) < 2, based on the induction hypothesis for S(k) and the inequality 1/(ak + 2 + 1) < 1/(ak + 1) mentioned earlier.
Since S(k + 1) < 2 and (2 - 1/(ak + 1) + 3/(ak + 1)) = (2ak + 3)/(ak + 1), we can conclude that:
S(k + 1) < (2ak + 3)/(ak + 1).
Therefore, the inequality holds for k + 1.
Step 4: Conclusion
Using mathematical induction, we have shown that if the inequality holds for n = 1 (base case) and for an arbitrary k (inductive hypothesis), it also holds for k + 1 (inductive step).
Therefore, the inequality 1/(a1 + 1) + 1/(a2 + 1) + ... + 1/(an + 1) < 2 is proven to be true for all n >= 1.
To prove the inequality, let's first try to find a pattern in the sequence (a_n). By substituting the equation a_n = a_(n - 1)^2 + a_(n - 1) into itself iteratively, we can compute the first few terms:
a_1 = 1/2
a_2 = (1/2)^2 + 1/2 = 1/4 + 1/2 = 3/4
a_3 = (3/4)^2 + 3/4 = 9/16 + 3/4 = 15/16
a_4 = (15/16)^2 + 15/16 = 225/256 + 15/16 = 255/256
From these computations, we can observe a pattern emerging. It seems that the terms are getting closer and closer to 1 as n increases.
To prove the inequality 1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_n + 1) < 2 for all n >= 1, we can use mathematical induction.
1. Base case: For n = 1, we have a_1 = 1/2. Substituting this value into the inequality, we get:
1/(1/2 + 1) = 1/(3/2) = 2/3 < 2
Thus, the inequality holds for the base case.
2. Inductive step: Assume that the inequality holds for some value k >= 1, i.e.,
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_k + 1) < 2.
We want to prove that the inequality also holds for k + 1, i.e.,
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_k + 1) + 1/(a_(k+1) + 1) < 2.
By using the inductive assumption, we have:
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_k + 1) < 2.
Adding 1/(a_(k+1) + 1) to both sides, we get:
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_k + 1) + 1/(a_(k+1) + 1) < 2 + 1/(a_(k+1) + 1).
We know that a_n is a non-negative sequence (since a_1, a_2, ... are all positive) and it approaches 1 as n goes to infinity. Therefore, a_(k+1) < 1.
Using this inequality, we can deduce that 1/(a_(k+1) + 1) > 1.
So, we have:
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_k + 1) + 1/(a_(k+1) + 1) < 2 + 1 = 3.
Therefore, the inductive step holds.
Based on the induction principle, we can conclude that the inequality 1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_n + 1) < 2 holds for all n >= 1.
To summarize, the key steps to solve this problem/proof are:
1. Observe the pattern in the given sequence.
2. Use induction to prove the inequality holds for the base case and the inductive step.
3. Use the properties of the sequence to establish the relationship between terms in the inequality.