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# Function

Consider a sequence a{n} given by a{1} =1/3, a{n+1} =a{n} + a{n} ^2. Let S=\sum{i=2}^2008 1/a_{I}, then find the value of [S] ? (where [ ] represent greatest integer function)

Note by Abhishek Pal
3 years, 3 months ago

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Compute the first few values

$$f(1) = \frac{1}{3}$$

$$f(2) = \frac{4}{9}$$

$$f(3) = \frac{52}{81}$$

$$f(4) = \frac{6916}{6561}$$

$$f(5) = \frac{93206932}{43046721}$$

Observe

$$\frac{1}{f(2)}+\frac{1}{f(3)}+\frac{1}{f(4)}+\frac{1}{f(5)} = \frac{121593159}{23301733} = B$$

The exact decimal representation of B is not important - it can be shown that $$5.2 < B < 5.3$$

Now lets try to put a bound on $$\sum\limits _{n=6}^{2008} \frac{1}{f(n)}$$

Let's try to calculate f(6):

$$f(6) = \frac{12699784969922596}{1853020188851841}$$

(Although the numbers are big this can all be done by hand in a reasonable amount of time)

It can be shown that $$f(6) > 6$$

From this, we see that $$f(7) > 6^2+6 = 42$$

What's more important is that for $$n>6$$, $$\frac{f(n)}{f(n-1)} > 2$$

And so, $$\frac{f(n-1)}{f(n)}<\frac{1}{2}$$

We conclude:

$$\frac{1}{f(6)} < \frac{1}{6}$$

$$\frac{1}{f(7)} < \frac{1}{12}$$

$$\frac{1}{f(8)} < \frac{1}{24}$$

$$\text{...}$$

$$\frac{1}{f(n+6)} < \frac{1}{2^n*6}$$

Now we have $$\sum\limits _{n=6}^{2008} \frac{1}{f(n)}=\sum\limits _{n=0}^{2002} \frac{1}{f(n+6)}<\sum\limits _{n=0}^{2002} \frac{1}{6 * 2^n}$$

We know that $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\text{...}+\frac{1}{2^n}<2$$ for every positive integer n.

And so, $$\sum\limits _{n=0}^{2002} \frac{1}{6 * 2^n}<\frac{1}{3}$$

And finally, $$\sum\limits _{n=6}^{2008} \frac{1}{f(n)} < \frac{1}{3}$$

Meaning that $$\sum\limits _{n=2}^{2008} \frac{1}{f(n)} < 5.3 + \frac{1}{3} < 6$$

The lower bound for $$\sum\limits _{n=2}^{2008} \frac{1}{f(n)}$$ remains $$5.2$$

This is sufficient to show that your sum is indeed between 5 and 6, and so your solution is 5. · 3 years, 2 months ago