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
TheDarkPrince
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 \[\sum_{j=1}^{N} \frac{1}{a_j + 1} = \frac{1}{2-1} - \frac{1}{a_N -1} = 1 - \frac{1}{a_N-1}<1 \qquad\blacksquare\]
leon.tyumen
28.10.2018 16:41
What mean RMO?
TheDarkPrince
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.
integrated_JRC
28.10.2018 17:01
$\textbf{\underline{Claim} :~}~a_n>n-1~~\forall~n\geqslant 1$
$\textbf{\underline{Proof of Claim} :~}~$
We can check the first few cases, viz. for $n=1(1)5$ , checking by hand, we can verify that our claimed result is true.
Now, for $k\geqslant 5$, we have \begin{align*}
(k-3)^2\geqslant 4>2 &\Leftrightarrow k^2-6k+7>0\\
&\Leftrightarrow k^2-4k+5>2k-2\\
&\Leftrightarrow \frac{k^2-4k+5}{2}>k-1\\
\end{align*}Now, according to our induction hypothesis, \begin{align*}
a_{k-1}>k-2 &\Leftrightarrow a_{k-1}^2+1>k^2-4k+5\\
&\Leftrightarrow a_k=\frac{a_{k-1}^2+1}{2}>\frac{k^2-4k+5}{2}\\
&\Leftrightarrow a_k>k-1\\
\end{align*}Q.E.D.
Now, back to the original problem, we can see $~a_n>0$ and $a_{n+1}=\frac{a_n^2+1}{2}>a_n$.
So, $a_1<a_2<a_3<a_4<\cdots$
Therefore, \begin{align*}
S_N&=\sum_{j=1}^{N} \frac{1}{a_j + 1}\\
&<\sum_{j=1}^{N} \frac{1}{a_1+ 1}\\
&=\frac{N}{a_N+1}~<~1~~\forall~n\in\mathbb{N}\\
\end{align*}
@below, sorry.
TheDarkPrince
28.10.2018 17:06
integrated_JRC wrote: \[\sum_{j=1}^{N} \frac{1}{a_1+ 1}=\frac{N}{a_N+1}\]
ShamimAkhtar212
15.02.2019 22:48
@TheDarkPrince How much you got in rmo?
TheDarkPrince
16.02.2019 08:37
ShamimAkhtar212 wrote: @TheDarkPrince How much you got in rmo? I was not even allowed to give RMO
hydrohelium
16.02.2019 09:08
What this cannot be true. Are you the boy who qualified for STEMS 2019 from CMI?
Delta0001
16.02.2019 11:29
@Above lol He is an INMO Merit Certificate holder
dram
08.07.2019 05:38
Does induction work here?
Math-wiz
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
Math-wiz
18.08.2019 08:48
The official solution can be seen here
Eulermathematics
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
melowmolly
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
Eulermathematics
14.01.2024 19:58
i know i am wrong but i want to know where am i wrong