Question

Question for Brainly Moderators, Brainly stars, best users. Sum the following series

$\sf \frac{5}{ {7}^{2} {3}^{2} } + \frac{9}{ {7}^{2} {11}^{2} } +\frac{13}{ {11}^{2} {15}^{2} } + - - - - \infty$

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Answer 1

$\large\underline{\sf{Solution-}}$

The given series is

$\rm :\longmapsto\:\sf \dfrac{5}{ {7}^{2} {3}^{2} } + \dfrac{9}{ {7}^{2} {11}^{2} } +\dfrac{13}{ {11}^{2} {15}^{2} } + - - - - \infty$

can be rewritten as

$\rm = \:\sf \dfrac{5}{ {3}^{2} {7}^{2} } + \dfrac{9}{ {7}^{2} {11}^{2} } +\dfrac{13}{ {11}^{2} {15}^{2} } + - - - - \infty$

Let assume that,

$\rm S_n = \:\sf \dfrac{5}{ {3}^{2} {7}^{2} } + \dfrac{9}{ {7}^{2} {11}^{2} } +\dfrac{13}{ {11}^{2} {15}^{2} } + - - - - n \: terms$

is further rewritten as by using nth term of an AP of the following AP series :-

$\rm :\longmapsto\:5,9,13, - - - , \: a_n = 4n + 1$

$\rm :\longmapsto\:3,7,11, - - - , \: a_n = 4n - 1$

$\rm :\longmapsto\:7,11, 15,- - - , \: a_n = 4n +3$

So, above series can be rewritten as

$\red{\rm S_n = \:\sf \dfrac{5}{ {3}^{2} {7}^{2} } + \dfrac{9}{ {7}^{2} {11}^{2} } +\dfrac{13}{ {11}^{2} {15}^{2} } + - - -\dfrac{4n + 1}{(4n - 1)^{2}(4n + 3)^{2}} }$

can be further rewritten as

$\red{\sf S_n = \dfrac{1}{8}\bigg[ \dfrac{40}{ {3}^{2} {7}^{2} } + \dfrac{72}{ {7}^{2} {11}^{2} } +\dfrac{104}{ {11}^{2} {15}^{2} } + - - -\dfrac{32n + 8}{(4n - 1)^{2} (4n + 3) ^{2} } \bigg]}$

$\red{\sf S_n = \dfrac{1}{8}\bigg[ \dfrac{49 - 9}{ {3}^{2} {7}^{2} } + \dfrac{121 - 49}{ {7}^{2} {11}^{2} } +\dfrac{225 - 121}{ {11}^{2} {15}^{2} } + - - -\dfrac{24n + 8n + 9 - 1}{(4n - 1)^{2} (4n + 3) ^{2} } \bigg]}$

$\red{\sf S_n = \dfrac{1}{8}\bigg[ \dfrac{ {7}^{2} - {3}^{2} }{ {3}^{2} {7}^{2} } + \dfrac{ {11}^{2} - {7}^{2} }{ {7}^{2} {11}^{2} } +\dfrac{ {15}^{2} - {11}^{2} }{ {11}^{2} {15}^{2} } + - - -\dfrac{ {(4n + 3)}^{2} - {(4n - 1)}^{2} }{(4n - 1)^{2} (4n + 3) ^{2} } \bigg]}$

$\red{ \sf \:S_n =\dfrac{1}{8}\bigg[\dfrac{1}{ {3}^{2} } - \dfrac{1}{ {7}^{2} } + \dfrac{1}{ {7}^{2} } - \dfrac{1}{ {11}^{2} } + \dfrac{1}{ {15}^{2} } - \dfrac{1}{ {11}^{2} } + - - + \dfrac{1}{ {(4n - 1)}^{2} } - \dfrac{1}{ {(4n + 3)}^{2} } \bigg] }$

$\red{ \sf \:S_n =\dfrac{1}{8}\bigg[\dfrac{1}{ {3}^{2} } - \dfrac{1}{ {(4n + 3)}^{2} } \bigg] }$

So,

$\red{ \sf \:\displaystyle\lim_{n \to \infty }S_n =\displaystyle\lim_{n \to \infty }\dfrac{1}{8}\bigg[\dfrac{1}{ {3}^{2} } - \dfrac{1}{ {(4n + 3)}^{2} } \bigg] }$

$\red{ \sf \:\displaystyle\lim_{n \to \infty }S_n =\dfrac{1}{8}\bigg[\dfrac{1}{ {3}^{2} } - 0 \bigg] }$

$\red{ \sf \:\displaystyle\lim_{n \to \infty }S_n =\dfrac{1}{72} }$

Hence,

$\:\boxed{ \tt{ \: \sf \dfrac{5}{ {7}^{2} {3}^{2} } + \dfrac{9}{ {7}^{2} {11}^{2} } +\dfrac{13}{ {11}^{2} {15}^{2} } + - - - - \infty = \frac{1}{72}}}$