calculate Σ_(n=6) ^∞ (1/(n^2 −25))


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Question Number 82289 by mathmax by abdo last updated on 19/Feb/20

$c a l c u l a t e \sum_{n = 6}^{\infty} \frac{1}{n^{2} - 25}$
Commented by mathmax by abdo last updated on 20/Feb/20
$let S_n =Σ_(k=6) ^n (1/(k^2 −25)) ⇒S_n =(1/(10))Σ_(k=6) ^n {(1/(k−5))−(1/(k+5))} ⇒ 10 S_n =Σ_(k=6) ^n (1/(k−5))−Σ_(k=6) ^n (1/(k+5)) we have Σ_(k=6) ^n (1/(k−5)) =_(k−5=p) Σ_(p=1) ^(n−5) (1/p) =H_(n−5) Σ_(k=6) ^n (1/(k+5)) =_(k+5=p) Σ_(p=11) ^(n+5) (1/p) =Σ_(p=1) ^(n+5) (1/p)−(1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6) +(1/7)+(1/8) +(1/9) +(1/(10)))=H_(n+5) −{(3/2)+(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99))} ⇒ 10S_n =H_(n−5) −H_(n+5) +(3/2) +(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99)) ∼ln(n−5)+γ +o((1/(n−5)))−γ−ln(n+5)−γ−o((1/(n+5))) +(3/2) +(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99)) ⇒10S_n =ln(((n−5)/(n+5)))+(3/2)+(7/(12))+((11)/(30)) +((15)/(56))+((19)/(99)) ⇒lim_(n→+∞) S_n =(1/(10))((3/2) +(7/(12)) +((11)/(30))+((15)/(56))+((19)/(99)))$
$l e t S_{n} = \sum_{k = 6}^{n} \frac{1}{k^{2} - 25} \Rightarrow S_{n} = \frac{1}{10} \sum_{k = 6}^{n} {\frac{1}{k - 5} - \frac{1}{k + 5}} \Rightarrow$ $10 S_{n} = \sum_{k = 6}^{n} \frac{1}{k - 5} - \sum_{k = 6}^{n} \frac{1}{k + 5} w e h a v e$ $\sum_{k = 6}^{n} \frac{1}{k - 5} =_{k - 5 = p} \sum_{p = 1}^{n - 5} \frac{1}{p} = H_{n - 5}$ $\sum_{k = 6}^{n} \frac{1}{k + 5} =_{k + 5 = p} \sum_{p = 11}^{n + 5} \frac{1}{p} = \sum_{p = 1}^{n + 5} \frac{1}{p} - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}$ $+ \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) = H_{n + 5} - {\frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{99}} \Rightarrow$ $10 S_{n} = H_{n - 5} - H_{n + 5} + \frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{99}$ $\sim l n (n - 5) + γ + o (\frac{1}{n - 5}) - γ - l n (n + 5) - γ - o (\frac{1}{n + 5})$ $+ \frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{99} \Rightarrow 10 S_{n} = l n (\frac{n - 5}{n + 5}) + \frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{99}$ $\Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{10} (\frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{99})$
Commented by mr W last updated on 20/Feb/20

$t y p o i n \frac{19}{99} . s h o u l d b e \frac{19}{90} .$
Commented by mathmax by abdo last updated on 20/Feb/20

$y e s s i r y o u a r e r i g h t t h a n k s$
Commented by mathmax by abdo last updated on 20/Feb/20

$f o r g i v e e r r o r o f t y p o c h a n g e 99 b y 90 \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = \frac{1}{10} (\frac{3}{2} + \frac{7}{12} + \frac{11}{30} + \frac{15}{56} + \frac{19}{90})$
	Answered by mr W last updated on 20/Feb/20
	$Σ_(n=6) ^∞ (1/(n^2 −25)) =(1/(10))Σ_(n=6) ^∞ ((1/(n−5))−(1/(n+5))) =(1/(10)){Σ_(n=6) ^∞ (1/(n−5))−Σ_(n=6) ^∞ (1/(n+5))} =(1/(10)){Σ_(n=1) ^∞ (1/n)−Σ_(n=10) ^∞ (1/(n+1))} =(1/(10)){Σ_(n=1) ^∞ (1/n)−Σ_(n=1) ^∞ (1/(n+1))+Σ_(n=1) ^9 (1/(n+1))} =(1/(10)){Σ_(n=1) ^∞ ((1/n)−(1/(n+1)))}+(1/(10))((1/2)+(1/3)+...+(1/(10))) =(1/(10))(1)+(1/(10))((1/2)+(1/3)+...+(1/(10))) =(1/(10))(1+(1/2)+(1/3)+...+(1/(10))) =((7381)/(25200))≈0.2929$
	$\underset{n = 6}{\sum^{\infty}} \frac{1}{n^{2} - 25}$ $= \frac{1}{10} \underset{n = 6}{\sum^{\infty}} (\frac{1}{n - 5} - \frac{1}{n + 5})$ $= \frac{1}{10} {\underset{n = 6}{\sum^{\infty}} \frac{1}{n - 5} - \underset{n = 6}{\sum^{\infty}} \frac{1}{n + 5}}$ $= \frac{1}{10} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 10}{\sum^{\infty}} \frac{1}{n + 1}}$ $= \frac{1}{10} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n + 1} + \underset{n = 1}{\sum^{9}} \frac{1}{n + 1}}$ $= \frac{1}{10} {\underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1})} + \frac{1}{10} (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{10})$ $= \frac{1}{10} (1) + \frac{1}{10} (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{10})$ $= \frac{1}{10} (1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{10})$ $= \frac{7381}{25200} \approx 0.2929$
	Commented by mathmax by abdo last updated on 20/Feb/20

	$t h a n k y o u s i r m r w .$
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