The question I am trying goes on like this

"Consider the sequence an given by $$ a_1 = \frac{1}{3}$$ $$ a_{n+1} = a_n^2 + a_n $$ Let $$ S = \frac{1}{a_2} + \frac{1}{a_3} + ... + \frac{1}{a_{2008}} $$ Then [S] is equal to _______. (where [.] represents the greatest integer function)

Please note that I haven't still learnt convergence and divergence of series, and so any help in formulating a solution not using the above mentioned topics would be appreciated.

I have tried calculating the values upto $a_5$ but have not been able to deduce a pattern.

  • 1
    $\begingroup$ Just to make sure: Did you state the problem correctly? Or do you mean this math.stackexchange.com/q/1453890/42969 ? $\endgroup$
    – Martin R
    Feb 19 '18 at 6:27
  • $\begingroup$ They are the exact wordings of the question $\endgroup$
    – saisanjeev
    Feb 19 '18 at 7:38
  • $\begingroup$ This has nothing to do with convergence. This is not a series, but a finite sum. It thus is equal to some real number. You don't need to think about convergence $\endgroup$
    – Yuriy S
    Feb 19 '18 at 8:10
  • 1
    $\begingroup$ Does it help any to notice that $1/(x^2+x)=(1/x)-(1/(x+1))$? $\endgroup$ Feb 19 '18 at 8:34
  • $\begingroup$ @gerry tried that but the 1/(a_i + 1) terms are creating some problems.(by a_i i mean the i'th term of the sequence $\endgroup$
    – saisanjeev
    Feb 19 '18 at 9:34

I think you don't even need a pattern at all to solve this. I solve this by using inequalities instead.

You can calculate that $a_2=\frac{4}{9}$; $a_3=\frac{52}{81}$; $a_4=\frac{6916}{6561}$; $a_5=\frac{93206932}{43046721}$, but it doesn't end here.

I use a calculator and know that $5.21<\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\frac{1}{a_5}<5.22$ (approximate range)

Also $a_5=\frac{93206932}{43046721}$ implies $1<2.165<a_5<2.166$, so $2.165^2+2.165<a_6<2.166^2+2.166$ or $6.852<a_6<6.858$, this continues for $53.801<a_7<53.891$ and $2948.348<a_8<2958.131$ and $8695704.277<a_9<8753497.145$.

Because of this and $\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\frac{1}{a_5}>5.21$, we conclude that $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_9}>5.21+\frac{1}{6.858}+\frac{1}{53.891}+\frac{1}{2958.131}+\frac{1}{8753497.145}>5.3747$

Note that all numbers are positive, so we can conclude that $S = \frac{1}{a_2} + \frac{1}{a_3} + ... + \frac{1}{a_{2008}}>5.3747$.

Also with $\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+\frac{1}{a_5}<5.22$, we conclude that $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_9}<5.22+\frac{1}{6.852}+\frac{1}{53.801}+\frac{1}{2948.348}+\frac{1}{8695704.277}<5.3849.$

Obviously, $\frac{1}{a_9}>\frac{1}{a_{10}}>\frac{1}{a_{11}}>...>\frac{1}{a_{2008}}$ because $a_9<a_{10}<a_{11}<...<a_{2008}$ and also have $a_9>8695704.277$, so $S = \frac{1}{a_2} + \frac{1}{a_3} + ... + \frac{1}{a_{2008}}<\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_9}+\frac{1}{a_9}\times 1999<5.3849+\frac{1999}{8695704.277}$ or $S<5.386$.

Combine with above, we will have $5.3747<S<5.386$, or $[S]=5$.

  • $\begingroup$ in the second last step, shouldn't be 1999/8695704.277 since you have written 1/a_9 * 1999, it will clearly not change the answer but still... and thanks a ton for the solid solution , requires a lot of patience for this inequality stuff for deducing the range $\endgroup$
    – saisanjeev
    Feb 19 '18 at 14:24
  • $\begingroup$ I edited and thanks for upvoting and accepting this answer $\endgroup$
    – user061703
    Feb 19 '18 at 14:45

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