2
$\begingroup$

Consider a sequence defined as, $$x_{n+1}=x_n^2+x_n\ \text{and}\ x_1=\frac{1}{3}$$ also $$P=\sum_{n=1}^{2018}\dfrac{1}{x_{n+1}}=\dfrac{1}{x_2}+\dfrac{1}{x_3}+\cdots+\dfrac{1}{x_{2019}}$$Find $[P]$. (where $[P]$ denotes largest integer less than or equal to $P$)

My tries:

In compact form, $$P=\sum_{n=1}^{2018}\dfrac{1}{x_n}-\dfrac{1}{x_n+1}$$

$$\dfrac{1}{x_{n+1}} =\dfrac{1}{x_n}-\dfrac{1}{x_n+1}\implies\displaystyle\sum_{n=1}^{2018}\dfrac{1}{x_n+\color{red}1}=\dfrac{1}{x_1}-\dfrac{1}{x_{2019}}$$but this gives me nothing. If somehow we can remove that $\color{red}1$, we are almost done, but how?

Then I try using some inequalities regarding $[.]$ as $$[x]+[y]\leq[x+y]$$

it just gave me lower bound as $3\leq[P]$.

Please help, I'm looking for some elegant answer, not simple (may not be simple with $2018$ terms) counting, Thanks!

$\endgroup$
  • $\begingroup$ You have 2019 in your question, which may suggest a 2019 competition. Which one is it, if any? $\endgroup$ – rtybase Feb 26 '19 at 21:00
3
$\begingroup$

Let us sum the first few terms:

$$ \sum_{n=1}^{5}\frac 1{x_n}=\frac 13+\frac{9}{4}+\frac{81}{52}+\frac{6561}{6916}+\frac{43046721}{93206932}=5.551\dots $$

Now we estimate $x_n$ from below:

$$ x_{n+1}=x_n^2+x_n>x_n^2\quad\Rightarrow\quad x_{n+1}>(x_1)^{2^n} $$

Then we can deduce that

$$ x_n>(x_6)^{2^{n-6}}=\left(\frac{12699784969922596}{1853020188851841}\right)^{2^{n-6}},\quad n\geq 7 $$

and estimate the sum

$$ \sum_{n=6}^{2018}\frac 1{x_n} <\sum_{n=6}^{2018}\frac 1{(x_6)^{2^{n-6}}} <\sum_{n=6}^{2018}\frac 1{(x_6)^{n-5}} <\sum_{n=6}^{\infty}\frac 1{(x_6)^{n-5}} =\frac{1}{x_6-1}=\frac{1853020188851841}{10846764781070755}=0.170\dots $$

So we get that

$$ 5<\sum_{n=1}^{2018}\frac{1}{x_n}<6\quad\Rightarrow\quad\left\lfloor \sum_{n=1}^{2018}\frac{1}{x_n}\right\rfloor=5. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.