Recursive Sequence Sum II

Algebra Level 4

Let a n a_n be a recursive function satisfying a n = a n − 1 − 1 n ( n − 1 ) a_n=a_{n-1}-\dfrac{1}{n(n-1)} for all positive integers n ≥ 2 n\geq 2 , and a 1 = 1 a_1=1 . What is the value of ∑ n = 1 1000 a n \displaystyle \sum_{n=1} ^{1000} a_n ? Round your answer to 5 decimal places.

Try Part I .


The answer is 7.48547.

Aaron Tsai by Aaron Tsai , 94, USA

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1 solution

Aaron Tsai
Aug 25, 2016

Relevant wiki: Harmonic Progression

We start by evaluating a n a_n for n = 2 , 3 , 4 n=2,3,4 .

a 2 = a 1 − 1 2 ( 2 − 1 ) = 1 − 1 2 = 1 2 a_2=\textcolor{#3D99F6}{a_1}-\dfrac{1}{2(2-1)}=\textcolor{#3D99F6}{1}-\dfrac{1}{2}=\dfrac{1}{2}

a 3 = a 2 − 1 3 ( 3 − 1 ) = 1 2 − 1 6 = 1 3 a_3=\textcolor{#3D99F6}{a_2}-\dfrac{1}{3(3-1)}=\textcolor{#3D99F6}{\dfrac{1}{2}}-\dfrac{1}{6}=\dfrac{1}{3}

a 4 = a 3 − 1 4 ( 4 − 1 ) = 1 3 − 1 12 = 1 4 a_4=\textcolor{#3D99F6}{a_3}-\dfrac{1}{4(4-1)}=\textcolor{#3D99F6}{\dfrac{1}{3}}-\dfrac{1}{12}=\dfrac{1}{4}

We note that a n − 1 \textcolor{#3D99F6}{a_{n-1}} is always 1 n − 1 \textcolor{#3D99F6}{\dfrac{1}{n-1}} . So, we can plug that into the function equation to get

a n = 1 n − 1 − 1 n ( n − 1 ) = 1 n − 1 + ( n − 1 ) − n n ( n − 1 ) = 1 n − 1 + ( 1 n − 1 n − 1 ) = 1 n a_n=\textcolor{#3D99F6}{\dfrac{1}{n-1}}-\dfrac{1}{n(n-1)}=\dfrac{1}{n-1}+\dfrac{(n-1)-n}{n(n-1)}=\dfrac{1}{n-1}+ \left ( \dfrac{1}{n} - \dfrac{1}{n-1} \right )= \dfrac{1}{n}

So, a n = 1 n a_n=\dfrac{1}{n} ; we are asked to find ∑ n = 1 1000 1 n \displaystyle \sum_{n=1} ^{1000} \dfrac{1}{n} , or the sum to 1000 1000 terms of the harmonic progression .

Using this approximation (where γ \gamma is the Euler-Mascheroni constant ) we find that

∑ n = 1 1000 1 n ≈ ln ⁡ 1000 + γ + 1 2 ( 1000 ) − 1 12 ( 1000 ) 2 + ⋅ ⋅ ⋅ ≈ 7.48547 \displaystyle \sum_{n=1} ^{1000} \dfrac{1}{n} \approx \ln{1000} + \gamma +\dfrac{1}{2(1000)} - \dfrac{1}{12(1000)^2} + \cdot \cdot \cdot \approx \boxed{7.48547}

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According to Jaume Oliver Lafont and also Wolfram Mathworld , a similar approximation that converges more quickly is: ∑ k = 1 n 1 k ≈ ln ⁡ ( n + 1 2 ) + γ + 1 24 ( n 2 ) − 1 24 ( n ) 3 + . . . ∴ ∑ k = 1 1000 1 k ≈ ln ⁡ ( 1000 + 1 2 ) + γ + 1 24 ( 100 0 2 ) − 1 24 ( 1000 ) 3 + . . . ≈ 7.48547 \begin{aligned} \displaystyle \sum_{k=1} ^{n} \dfrac{1}{k} & \approx \ln{(n+\frac12)} + \gamma +\dfrac{1}{24(n^2)} - \dfrac{1}{24(n)^3}+...\\ \therefore \displaystyle \sum_{k=1} ^{1000} \dfrac{1}{k} & \approx \ln{(1000+\frac12)} + \gamma +\dfrac{1}{24(1000^2)} - \dfrac{1}{24(1000)^3}+... \approx \boxed{7.48547} \end{aligned} This converges to 5 decimal places correctly with just ln ⁡ ( 1000.5 ) + γ \ln{(1000.5)}+\gamma

Richard Costen - 4 years, 9 months ago

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