The discussion revolves around the integer part of the sum $$\sum_{n=1}^{2001}\dfrac{1}{a_n+1}$$ where the sequence is defined by the initial term $$a_1=\dfrac{1}{3}$$ and the recursive relation $$a_{n+1}=a_n^2+a_n$$ for $$n=1,2,\ldots,2000$$. Participants are exploring the properties of the sequence and the implications for the sum.
There is no consensus on the exact value of the integer part of the sum, and multiple competing views on how to approach the problem remain. Participants are exploring various methods and reasoning without reaching a definitive conclusion.
Participants note potential limitations in their approaches, including assumptions about the growth rate of the sequence and the behavior of the sum as $$n$$ increases. Some mathematical steps remain unresolved, particularly in relation to the convergence of the series.
This discussion may be useful for those interested in sequences, series, and mathematical reasoning, particularly in the context of recursive definitions and their implications for sums.
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.