# Sequence summation: $x_{n+1}=x_n^2+x_n$ and $x_1=\frac{1}{3}$

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!

• You have 2019 in your question, which may suggest a 2019 competition. Which one is it, if any? – 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

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