Define a sequence $\{a_n\}_{n\geq 1}$ of real numbers by \[a_1=2,\qquad a_{n+1} = \frac{a_n^2+1}{2}, \text{ for } n\geq 1.\]Prove that \[\sum_{j=1}^{N} \frac{1}{a_j + 1} < 1\]for every natural number $N$.
Problem
Source: RMO 2018 P6
Tags: algebra
28.10.2018 16:35
Solution: \begin{align*} \frac{1}{a_k + 1} = \frac{a_k-1}{a_k^2 - 1}&= \frac{a_k+1-2}{a_k^2 - 1}\\ &= \frac{1}{a_k - 1} - \frac{2}{a_k^2 - 1} = \frac{1}{a_k-1} - \frac{1}{a_{k+1} -1}. \end{align*} Thus \[\sum_{j=1}^{N} \frac{1}{a_j + 1} = \frac{1}{2-1} - \frac{1}{a_N -1} = 1 - \frac{1}{a_N-1}<1 \qquad\blacksquare\]
28.10.2018 16:41
What mean RMO?
28.10.2018 16:44
leon.tyumen wrote: What mean RMO? Regional Maths Olympiad. Second stage of olympiad in India. RMO happened on 7th Oct and today this year.
28.10.2018 17:01
@below, sorry.
28.10.2018 17:06
integrated_JRC wrote: \[\sum_{j=1}^{N} \frac{1}{a_1+ 1}=\frac{N}{a_N+1}\]
15.02.2019 22:48
@TheDarkPrince How much you got in rmo?
16.02.2019 08:37
ShamimAkhtar212 wrote: @TheDarkPrince How much you got in rmo? I was not even allowed to give RMO
16.02.2019 09:08
What this cannot be true. Are you the boy who qualified for STEMS 2019 from CMI?
16.02.2019 11:29
@Above lol He is an INMO Merit Certificate holder
08.07.2019 05:38
Does induction work here?
15.08.2019 13:12
dram wrote: Does induction work here? The official solution to the problem involved induction IIRC. The HBCSE website has some sort of problem I guess, do I am not able to access that file
18.08.2019 08:48
The official solution can be seen here
06.01.2024 16:45
is this problem wrong bcoz a_n+1-a_n = (a_n-1)^2/2 then summing that from n =1 to infinity we have sum of (a_n-1)^2 = -4
07.01.2024 12:26
Eulermathematics wrote: is this problem wrong bcoz a_n+1-a_n = (a_n-1)^2/2 then summing that from n =1 to infinity we have sum of (a_n-1)^2 = -4 this sequence has a limit of infinity so you are wrong
14.01.2024 19:58
i know i am wrong but i want to know where am i wrong