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Evaluate $X$ in its simplest form then find the sum of all digits of $X$. Where $X$ is given as
$$X=\sqrt{2008+2007\sqrt{2008+2007\sqrt{2008+2007\sqrt{\cdots}}}}$$

I tried the following method:

\begin{align} X^2 &={2008+2007\sqrt{2008+2007\sqrt{2008+2007\sqrt{\cdots}}}} \tag1\\ X^2 &={2008+2007X} \tag2\\ X^2-2008-2007X &=0 \tag3\\ (X+1)(X-2008) &=0 \tag4\\ X &=-1, 2008 \tag5 \end{align}

But now we have two solutions for the sum of digits.

Where did I go wrong? Is the method correct?

Really appreciate the help, thank you!!

Blue
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KnightRiderDutt
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1 Answers1

2

The expressions you are dealing with are all positive, so only the positive solution makes sense. But before that, you need to know that there is a solution.

For this, consider the recursive sequence $$ a_1=1,\qquad a_{n+1}=\sqrt{2008+2007 a_n}. $$

  • $a_n\leq2008$ for all $n$. By induction, $0<a_1=1\leq 2008$, and if $0<a_n\leq2008$ then $$ a_{n+1}=\sqrt{2008+2007 a_n}\leq\sqrt{2008+2007\times 2008}=2008. $$

  • The sequence is monotone. Indeed, $$ a_{n+1}=\sqrt{2008+2007 a_n}\geq\sqrt{a_n+2007a_n}=\sqrt{2008a_n}\geq\sqrt{a_n^2}=a_n. $$

The two conditions together imply that $L=\lim_na_n$ exists. Taking limit in $$ a_{n+1}=\sqrt{2008+2007a_n} $$ we get $$ L=\sqrt{2008+2007L}. $$ Now you can find that $L=2008$ as you did.

Martin Argerami
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