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Evaluate the sum \(\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\)

 Mar 5, 2019
 #1
avatar
+1

sumfor(n, 1, infinity, 6/((2*n + 1)^2 - 1)) = 3 / 2

 Mar 5, 2019
 #2
avatar+23064 
+2

help

\(\large{\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots}\)

 

\(\begin{array}{|rcll|} \hline && \dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots \\\\ &=& \dfrac{6}{(3-1)(3+1)}+\dfrac{6}{(5-1)(5+1)}+\dfrac{6}{(7-1)(7+1)}+\dfrac{6}{(9-1)(9+1)}+\cdots \\\\ &=& \dfrac{6}{2\cdot 4}+\dfrac{6}{4\cdot 6}+\dfrac{6}{6\cdot 8}+\dfrac{6}{8\cdot 10}+\cdots \\\\ &=& 6\left( \dfrac{1}{2\cdot 4}+\dfrac{1}{4\cdot 6}+\dfrac{1}{6\cdot 8}+\dfrac{1}{8\cdot 10}+\cdots \right) \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{2k(2k+2)} \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{4k(k+1)} \\\\ &=& \dfrac{6}{4}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \quad | \quad \boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} } \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{k+1} \right) \\\\ &=& \dfrac{3}{2}\cdot \left[\sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot (1+0) \\\\ &=& \dfrac{3}{2} \\ \hline \end{array}\)

 

laugh

 Mar 5, 2019
 #3
avatar+103120 
+1

Note that

3^2 - 1 =  8 = 2*4

5^2 - 1 = 24 = 4*6

7^2 - 1 = 48 =  6*8

 

So  we have  for n = 1 to infinity

 

6 *    1 / [ 2n * (2n + 2) ]  =  6 *   1 / [ 4 (n)(n+ 1) ]   =  (3/2) * 1/ [ n * (n + 1) ]

 

Note that     1  / [ n * (n + 1) ]   can be written as    1 /  n - 1/{n + 1]

 

So we have

 

(3/2)  [ ( 1 - 1/2 ) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ...... ]    

 

All the terms cancel  except  for the first term , 1,   and the last term , -1/(n + 1)

 

And when n is large  - 1 (n + 1) → 0

 

So....the sum is

 

(3/2) (1)  =

 

3/2

 

 

cool cool cool

 Mar 5, 2019

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