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# Please provide me a formal approach

I tried this question some days ago at a discussion but could not get a formal method : Consider a sequence $$a_{n}$$ given by $$a_{1} =\frac{1}{3}, a_{n+1} =a_{n} + a_{n} ^2$$. Let $$S=\displaystyle \sum_{r=2}^{2008} \frac{1}{a_{r}}$$, then find the value of $$\lfloor S\rfloor$$?

Note by Kushagraa Aggarwal
4 years, 6 months ago

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By Induction we can prove that for $$n>6, a_n > n^2$$ . $$\Rightarrow \frac {1}{a_n} < \frac {1}{n^2}$$

It's obvious that all $$a_i$$ are positive

$$\displaystyle \sum_{i=2}^6 \frac {1}{a_i} < \displaystyle \sum_{i=2}^{2008} \frac {1}{a_i}$$

$$\displaystyle \sum_{i=2}^6 \frac {1}{a_i} < \displaystyle \sum_{i=2}^6 \frac {1}{a_i} + \displaystyle \sum_{i=7}^{2008} \frac {1}{a_i}$$

$$\displaystyle \sum_{i=2}^6 \frac {1}{a_i} < \displaystyle \sum_{i=2}^6 \frac {1}{a_i} + \displaystyle \sum_{i=7}^{2008} \frac {1}{a_i} < \displaystyle \sum_{i=2}^6 \frac {1}{a_i} + \displaystyle \sum_{i=7}^\infty \frac {1}{n^2}$$

$$\large 5.36... < S < 5.36... + ( \frac {\pi^2}{6} - \frac {1}{1^2} - \frac {1}{2^2} - \frac {1}{3^2} - \frac {1}{4^2} - \frac {1}{5^2} - \frac {1}{6^2} )$$

$$\large 5.36... < S < 5.36... + ( \frac {\pi^2}{6} - \frac {5369}{3600} ) < 5.36... + ( \frac {(22/7)^2}{6} - \frac {5369}{3600} )$$

$$\large 5.36... < S < 5.51...$$

Hence, $$\lfloor S \rfloor = 5$$

- 4 years, 6 months ago

Using the recursion, we have: $$a_2 = \dfrac{4}{9}$$, ... , $$a_8 \approx 8.7110 \times 10^6$$. So, $$5.3830 < \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} < 5.3831$$.

Since $$a_n$$ is increasing, we have $$a_r > a_8 > 0$$ for $$r > 8$$. So, $$0 < \dfrac{1}{a_r} < \dfrac{1}{a_8} < 1.1480 \times 10^{-7}$$ for all $$r > 8$$.

Then, $$5 < 5.3830 < \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r}$$ $$< \displaystyle\sum_{r = 2}^{2008}\dfrac{1}{a_r}$$ $$= \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} + \displaystyle\sum_{r = 9}^{2008}\dfrac{1}{a_r}$$ $$< \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} +\dfrac{2000}{a_8} < 5.3834 < 6$$.

Therefore, $$5 < S < 6$$, and thus, $$\left\lfloor S \right\rfloor = 5$$.

To "formalize" this, replace all the approximations with exact fractions, then use similar bounds.

- 4 years, 6 months ago