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 \(\lfloor S\rfloor\)?

#Proofs #MathProblem #Math

Note by Kushagraa Aggarwal
7 years, 10 months ago

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Comments

By Induction we can prove that for n>6,an>n2n>6, a_n > n^2 . ⇒1an<1n2 \Rightarrow \frac {1}{a_n} < \frac {1}{n^2}

It's obvious that all aia_i are positive

∑i=261ai<∑i=220081ai \displaystyle \sum_{i=2}^6 \frac {1}{a_i} < \displaystyle \sum_{i=2}^{2008} \frac {1}{a_i}

∑i=261ai<∑i=261ai+∑i=720081ai \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}

∑i=261ai<∑i=261ai+∑i=720081ai<∑i=261ai+∑i=7∞1n2 \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}

5.36...<S<5.36...+(π26−112−122−132−142−152−162) \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} )

5.36...<S<5.36...+(π26−53693600)<5.36...+((22/7)26−53693600) \large 5.36... < S < 5.36... + ( \frac {\pi^2}{6} - \frac {5369}{3600} ) < 5.36... + ( \frac {(22/7)^2}{6} - \frac {5369}{3600} )

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

Hence, ⌊S⌋=5 \lfloor S \rfloor = 5

Pi Han Goh - 7 years, 9 months ago

Using the recursion, we have: a2=49a_2 = \dfrac{4}{9}, ... , a8≈8.7110×106a_8 \approx 8.7110 \times 10^6. So, 5.3830<∑r=281ar<5.38315.3830 < \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} < 5.3831.

Since ana_n is increasing, we have ar>a8>0a_r > a_8 > 0 for r>8r > 8. So, 0<1ar<1a8<1.1480×10−70 < \dfrac{1}{a_r} < \dfrac{1}{a_8} < 1.1480 \times 10^{-7} for all r>8r > 8.

Then, 5<5.3830<∑r=281ar5 < 5.3830 < \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} <∑r=220081ar< \displaystyle\sum_{r = 2}^{2008}\dfrac{1}{a_r} =∑r=281ar+∑r=920081ar= \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} + \displaystyle\sum_{r = 9}^{2008}\dfrac{1}{a_r} <∑r=281ar+2000a8<5.3834<6< \displaystyle\sum_{r = 2}^{8}\dfrac{1}{a_r} +\dfrac{2000}{a_8} < 5.3834 < 6.

Therefore, 5<S<65 < S < 6, and thus, ⌊S⌋=5\left\lfloor S \right\rfloor = 5.

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

Jimmy Kariznov - 7 years, 9 months ago
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