+0

# Sequences Help Fast Thank you!

+1
275
6
+49

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

Feb 20, 2019

#2
+24983
+4

Evaluate the sum

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

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

Feb 21, 2019
edited by heureka  Feb 22, 2019
edited by heureka  Feb 22, 2019

#1
+6187
+1

$$\text{A little toying with this and lookup at OEIS show that this sum is equal to}\\ \dfrac 3 4\sum \limits_{k=1}^\infty~\dbinom{k+1}{2}^{-1} = \dfrac 3 2$$

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Feb 20, 2019
#2
+24983
+4

Evaluate the sum

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

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

heureka Feb 21, 2019
edited by heureka  Feb 22, 2019
edited by heureka  Feb 22, 2019
#3
+111329
+1

Very nice, Heureka......I wondered if there was a way to determine this "by hand"

CPhill  Feb 21, 2019
#4
+1904
+2

Hello Heureka,

In your solution, one of the (k) factors seems to have randomly evaporated:

$$\dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{\underbrace {k}_{evaporated?}(k+1)} \quad | \quad \boxed{\dfrac{1}{k+1} = \dfrac{1}{k} - \dfrac{1}{k+1} } \\\\ \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{k+1} \right) \\\\$$

I thought you would like to fix this fuckup

GA

GingerAle  Feb 21, 2019
#5
+24983
+3

Sorry GA,

I corrected my typo:

$$\boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} }$$

heureka  Feb 22, 2019
edited by heureka  Feb 22, 2019
#6
+24983
+3

Thank you CPhill,

heureka  Feb 22, 2019