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Compute \(\sum_{n = 1}^\infty \frac{2n - 1}{n(n + 1)(n + 2)}\)

 Feb 25, 2019

Best Answer 

 #1
avatar+17332 
+2

I went to Symbolab for this one:

 

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%5Cinfty%20%7D%5Cleft(%5Cfrac%7B%5Cleft(2x-1%5Cright)%7D%7B%5Cleft(x%5Cright)%5Cleft(x%2B1%5Cright)%5Cleft(x%2B2%5Cright)%7D%5Cright)

 Feb 25, 2019
 #1
avatar+17332 
+2
Best Answer

I went to Symbolab for this one:

 

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%5Cinfty%20%7D%5Cleft(%5Cfrac%7B%5Cleft(2x-1%5Cright)%7D%7B%5Cleft(x%5Cright)%5Cleft(x%2B1%5Cright)%5Cleft(x%2B2%5Cright)%7D%5Cright)

ElectricPavlov Feb 25, 2019
 #2
avatar
0

The link doesn't work on edge

Guest Feb 25, 2019
edited by Guest  Feb 25, 2019
 #3
avatar+98168 
+1

 

One of our members, heureka, may be able to present the solution in detail

 

 

cool cool cool

 Feb 25, 2019
 #4
avatar+99329 
+1

Wolfram|Alpha says to do it using the comparison test.

 

Here is a video on the comparison test.

 

Note: I have not wached the video but Khan Academy produces many great video clips.

 

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/comparison-test-convergence

 Feb 25, 2019
 #5
avatar
+1

It is the sum of this sequence:

 

1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6) +...............+ (2n - 1) / n(n+1(n+2) = 3 / 4

 Feb 25, 2019
 #6
avatar+99329 
0

Yes, we already know that.

Melody  Feb 25, 2019
 #7
avatar+21848 
+2

Compute

\(\mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}}}\)

 

\(\begin{array}{lcll} \mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\ \begin{array}{|lcll|} \hline s_n = \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

 

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

 

We rearrange:
\(\begin{array}{|rcll|} \hline \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} &=&\left(\dfrac{2n - 1}{n}\right)\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \left(\dfrac{2n - 1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \left(2-\dfrac{1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{n(n+1)}+\dfrac{1}{n(n+2)} \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)+ \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{n}+\dfrac{1}{2}\cdot \dfrac{1}{n}+\dfrac{2}{n+1}+\dfrac{1}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{2}\cdot \left(\dfrac{1}{n+2}\right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-\dfrac{5}{2(n+2)} \quad | \quad \dfrac{1}{2(n+2)} = \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-5\cdot \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1} -\dfrac{5}{2n} + \dfrac{5}{n(n+2)} \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{n(n+2)} \quad | \quad \dfrac{1}{n(n+2)} = \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + 5\cdot \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} \\\\ \mathbf{\dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} } \\ \hline \end{array}\)

 

Telescoping series:

\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{-3} &\color{red}+& \color{red}\dfrac{3}{2} &+& \mathbf{\dfrac{5}{2}\cdot \dfrac{1}{1}} &\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{3} \\\\ &\color{red}-& \color{red}\dfrac{3}{2} &\color{red}+& \color{red}\dfrac{3}{3} &+&\mathbf{\dfrac{5}{2}\cdot \dfrac{1}{2}}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{4} \\\\ &\color{red}-& \color{red}\dfrac{3}{3} &\color{red}+& \color{red}\dfrac{3}{4} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{3}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{5} \\\\ &\color{red}-& \color{red}\dfrac{3}{4} &\color{red}+& \color{red}\dfrac{3}{5} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{4}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{6} \\\\ &\color{red}-& \color{red}\dfrac{3}{5} &\color{red}+& \color{red}\dfrac{3}{6} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{5}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{7} \\\\ &\color{red}-& \color{red}\dfrac{3}{6} &\color{red}+& \color{red}\dfrac{3}{7} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{6}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{8} \\\\ && \ldots \\\\ &\color{red}-& \color{red}\dfrac{3}{n-3} &\color{red}+& \color{red}\dfrac{3}{n-2} &\color{green}+&\color{green}\dfrac{5}{2(n-3)}&\color{green}-& \color{green}\dfrac{5}{2(n-1)} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-2} &\color{red}+& \color{red}\dfrac{3}{n-1} &\color{green}+&\color{green}\dfrac{5}{2(n-2)}&\color{green}-& \color{green}\dfrac{5}{2n} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-1} &\color{red}+& \color{red}\dfrac{3}{n} &\color{green}+&\color{green}\dfrac{5}{2(n-1)}&-& \mathbf{\dfrac{5}{2(n+1)}} \\\\ &\color{red}-& \color{red}\dfrac{3}{n} &+& \mathbf{\dfrac{3}{n+1}} &\color{green}+&\color{green}\dfrac{5}{2n}&-& \mathbf{\dfrac{5}{2(n+2)}} \\ \hline \end{array}\)

 

The colored terms shorten out

 

So \(s_n\) is, we have all black terms left :
\(\begin{array}{|rcll|} \hline s_n &=& -3 + \dfrac{5}{2} + \dfrac{5}{2}\cdot \dfrac{1}{2} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{3}{4} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} } \\ \hline \end{array} \)

 

\( \lim \limits_{n\to \infty} { \dfrac{5}{2(n+1)}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{3}{n+1} } = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{5}{2(n+2)} } = 0\)

 

\(\begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{3}{4} - 0 + 0 -0 \\\\ &=& \dfrac{3}{4} \\ \hline \end{array}\)

 

\(\begin{array}{lcll} \mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}} = \dfrac{3}{4} } \\ \end{array}\\ \)

 

laugh

 Feb 26, 2019
 #8
avatar+98168 
0

Wow!!!!....impressive, Heureka!!!

 

 

cool cool cool

CPhill  Feb 26, 2019

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