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A sequence is defined as follows: $a_1=\frac{1}{2}$ and $a_{n+1}=a_n^2+a_n$

If $$S=\frac{1}{a_1+1}+\frac{1}{a_2+1}+\frac{1}{a_3+1}+.....+\frac{1}{a_{100}+1}$$

Find $\lfloor S \rfloor$.

Now using $a_{n+1}=a_n^2+a_n$, I generated telescopic series to get $S=2-\frac{1}{a_{101}}$ but I am not able to find $\frac{1}{a_{101}}$

Could someone help me with this?

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You don't need actually ${1\over a_{101}}$. Since $a_{n+1}-a_n = a_n^2>0$ and $a_3>1$ we have $$0<{1\over a_{101}}<1$$ so

$$ 1 <2-{1\over a_{101}}<2 \Longrightarrow \lfloor S \rfloor = 1$$

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  • $\begingroup$ Note that $a_n<1$, for $n<3$. More verification is needed. $\endgroup$ – Ng Chung Tak Feb 12 '18 at 17:37
  • $\begingroup$ Why? How does this affect on $a_{101}$? $\endgroup$ – Aqua Feb 12 '18 at 17:40
  • $\begingroup$ It depends on initial condition. If $a_1$ belongs to the Mandelbrot Set, then $|a_{n}|<2 \:, \: \forall n\in \mathbb{N}$. $\endgroup$ – Ng Chung Tak Feb 12 '18 at 17:50
  • $\begingroup$ But $a_1 =1/2$. So $a_2 = 3/4$ and $a_3= 21/16>1$ so $a_n>1$ for all $n>2$. $\endgroup$ – Aqua Feb 12 '18 at 17:51
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    $\begingroup$ @ChristianF So we cannot calculate $S$ in few steps ? I guess that is why question asked for integral part of $S$. $\endgroup$ – Mathematics Feb 12 '18 at 17:58

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