*{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)

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TopNewestYour answer is 5.

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.

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according to me it is correct but according to book it is incorrect.............................the answer is.....2

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