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.
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.
2 answers
Thank you. I'll get to work on that immiediately.
3 answers