hint:Albert said:$$a\,\, sequence:a_1=\dfrac {1}{3}\,\, and \,\, \,\, a_{n+1}=a_n^2+a_n---(1),\,\, n=1,2,3,----2000\\
find \,\, the \,\, integer \,\, part\,\, of :\sum_{n=1}^{2001}\dfrac {1}{a_{n}+1}$$
lfdahl said:My solution(s):
A. First a quantitative approach.We can rewrite the sum, by noting that:
\[a_{n+1} = a_n^2+a_n \Rightarrow a_{n+1}+1 = a_n^2+a_n+1\Rightarrow \frac{1}{a_{n+1}+1} = \frac{1}{a_n^2+a_n+1}\]
Thus:
\[\sum_{n=1}^{2001}\frac{1}{a_n+1} = \frac{1}{a_1+1}+\sum_{n=1}^{2000}\frac{1}{a_n^2+a_n+1} \]
For $n \ge 4$ the $a_n$ are greater than 1, and grow in power of $2$, so the sum converges rapidly. Please cf. the table below.The first $9$ terms:
\[\begin{matrix} n & a_n & \frac{1}{a_n+1} & \sum_{i=1}^{n}\frac{1}{a_i+1}\\ 1 & 1/3 \approx 0.333 & 3/4 =0.75 & 3/4 = 0.75\\ 2& 4/9 \approx 0.444 & 9/13\approx 0.692 & 75/52 \approx 1.442\\ 3& 52/81 \approx 0.642& 81/133 \approx 0.609 & \frac{14187}{6916} \approx 2.051\\ 4& \frac{6916}{6561} \approx 1.054 & \frac{6561}{13477}\approx 0.487 & \approx 2.538\\ 5& \approx 2.165 & \approx 0.316 & \approx 2.854\\ 6&\approx 6.854 & \approx 0.127 & \approx 2.981\\ 7& \approx 53.82 & \approx 0.018 & \approx 2.999\\ 8& \approx 2951 & \approx 0.0003 & \approx 3.000\\ 9& \approx 8710990 & \approx 0.0000001 & \approx 3.000 \end{matrix}\]Thus the integer part of the sum is $3$.B. Then there is a qualitative approach:
Rewriting the recursive relation:
\[a_{n+1} = a_n^2+a_n \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_n(a_n+1)} = \frac{1}{a_n}-\frac{1}{a_n+1} \Rightarrow \frac{1}{a_n+1} = \frac{1}{a_n}-\frac{1}{a_{n+1}}\]
So we have a telescoping sum:
\[\sum_{n = 1}^{2001} \frac{1}{a_n+1}= \sum_{n = 1}^{2001}\left ( \frac{1}{a_n} -\frac{1}{a_{n+1}}\right ) \\\\ =\left ( \frac{1}{a_1}-\frac{1}{a_2} \right )+\left (\frac{1}{a_2}-\frac{1}{a_3} \right )+ \left (\frac{1}{a_3}-\frac{1}{a_4} \right )+...+ \left (\frac{1}{a_{2001}}-\frac{1}{a_{2002}} \right )=\frac{1}{a_1}-\frac{1}{a_{2002}} = \frac{1}{a_1} = 3.\]
From the table, we know, that $a_n$ grows rapidly above $1$ for $n >4$.
Hence, the last term $\frac{1}{a_{2002}}$ can be ignored.
Albert said:$2<\dfrac{1}{a_1}-\dfrac{1}{a_{2002}}<3$
so the integer part should be 2
Neatlfdahl said:My solution(s):
A. First a quantitative approach.We can rewrite the sum, by noting that:
\[a_{n+1} = a_n^2+a_n \Rightarrow a_{n+1}+1 = a_n^2+a_n+1\Rightarrow \frac{1}{a_{n+1}+1} = \frac{1}{a_n^2+a_n+1}\]
Thus:
\[\sum_{n=1}^{2001}\frac{1}{a_n+1} = \frac{1}{a_1+1}+\sum_{n=1}^{2000}\frac{1}{a_n^2+a_n+1} \]
For $n \ge 4$ the $a_n$ are greater than 1, and grow in power of $2$, so the sum converges rapidly. Please cf. the table below.The first $9$ terms:
\[\begin{matrix} n & a_n & \frac{1}{a_n+1} & \sum_{i=1}^{n}\frac{1}{a_i+1}\\ 1 & 1/3 \approx 0.333 & 3/4 =0.75 & 3/4 = 0.75\\ 2& 4/9 \approx 0.444 & 9/13\approx 0.692 & 75/52 \approx 1.442\\ 3& 52/81 \approx 0.642& 81/133 \approx 0.609 & \frac{14187}{6916} \approx 2.051\\ 4& \frac{6916}{6561} \approx 1.054 & \frac{6561}{13477}\approx 0.487 & \approx 2.538\\ 5& \approx 2.165 & \approx 0.316 & \approx 2.854\\ 6&\approx 6.854 & \approx 0.127 & \approx 2.981\\ 7& \approx 53.82 & \approx 0.018 & \approx 2.999\\ 8& \approx 2951 & \approx 0.0003 & \approx 3.000\\ 9& \approx 8710990 & \approx 0.0000001 & \approx 3.000 \end{matrix}\]Thus the integer part of the sum is $3$.B. Then there is a qualitative approach:
Rewriting the recursive relation:
\[a_{n+1} = a_n^2+a_n \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_n(a_n+1)} = \frac{1}{a_n}-\frac{1}{a_n+1} \Rightarrow \frac{1}{a_n+1} = \frac{1}{a_n}-\frac{1}{a_{n+1}}\]
So we have a telescoping sum:
\[\sum_{n = 1}^{2001} \frac{1}{a_n+1}= \sum_{n = 1}^{2001}\left ( \frac{1}{a_n} -\frac{1}{a_{n+1}}\right ) \\\\ =\left ( \frac{1}{a_1}-\frac{1}{a_2} \right )+\left (\frac{1}{a_2}-\frac{1}{a_3} \right )+ \left (\frac{1}{a_3}-\frac{1}{a_4} \right )+...+ \left (\frac{1}{a_{2001}}-\frac{1}{a_{2002}} \right )=\frac{1}{a_1}-\frac{1}{a_{2002}} = \frac{1}{a_1} = 3.\]
From the table, we know, that $a_n$ grows rapidly above $1$ for $n >4$.
Hence, the last term $\frac{1}{a_{2002}}$ can be ignored.