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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 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, \( \lfloor S \rfloor = 5 \) Pi Han Goh · 4 years ago

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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. Jimmy Kariznov · 4 years ago

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