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!

  • $\begingroup$ 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

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

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


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