MHB Integer Part of $$\sum_{n=1}^{2001}\dfrac{1}{a_n+1}$$

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The sequence defined by \( a_1 = \frac{1}{3} \) and \( a_{n+1} = a_n^2 + a_n \) grows rapidly, leading to large values for \( a_n \) as \( n \) increases. Consequently, the terms \( \frac{1}{a_n + 1} \) approach zero for larger \( n \). The sum \( \sum_{n=1}^{2001} \frac{1}{a_n + 1} \) converges to a finite value, primarily influenced by the initial terms of the sequence. The integer part of this sum can be calculated by evaluating the first few terms, which dominate the overall sum. Thus, the integer part of the sum is determined to be 1.
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$$a\,\, sequence:a_1=\dfrac {1}{3}\,\, and \,\, \,\, a_{n+1}=a_n^2+a_n,\,\, n=1,2,3,----2000\\
find \,\, the \,\, integer \,\, part\,\, of :\sum_{n=1}^{2001}\dfrac {1}{a_{n}+1}$$
 
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Albert said:
$$a\,\, sequence:a_1=\dfrac {1}{3}\,\, and \,\, \,\, a_{n+1}=a_n^2+a_n---(1),\,\, n=1,2,3,----2000\\
find \,\, the \,\, integer \,\, part\,\, of :\sum_{n=1}^{2001}\dfrac {1}{a_{n}+1}$$
hint:
take reciprocals of both sides of (1)
and note: $a_{2001}>a_{2000}>-------->a_4>1>a_3>a_2>a_1=\dfrac {1}{3}$
 
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.
 
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.
$2<\dfrac{1}{a_1}-\dfrac{1}{a_{2002}}<3$
so the integer part should be 2
 
Albert said:
$2<\dfrac{1}{a_1}-\dfrac{1}{a_{2002}}<3$
so the integer part should be 2
I´m sorry for my wrong conclusion. The integer part must be $2$. This is not obvious in the quantitative approach, because, I´ve made a rounding error. Thankyou for a most challenging puzzle, Albert!
 
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.
Neat
 
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