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Question Number 114023 by Khalmohmmad last updated on 16/Sep/20

Σ_(n=1) ^∝ (1/(n^2 −1))=?

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\propto} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{1}}=? \\ $$

Commented by Dwaipayan Shikari last updated on 16/Sep/20

Diverges first term (1/(n^2 −1))→∞( n=1 ) or undefined

$${Diverges} \\ $$$${first}\:{term}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\rightarrow\infty\left(\:\:{n}=\mathrm{1}\:\:\right)\:{or}\:{undefined} \\ $$

Answered by mr W last updated on 16/Sep/20

Σ_(n=2) ^∝ (1/(n^2 −1)) =(1/2)Σ_(n=2) ^∝ ((1/(n−1))−(1/(n+1))) =(1/2)((1/1)−(1/3)+(1/2)−(1/4)+(1/3)−(1/5)+...) =(1/2)×(3/2) =(3/4)

$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\propto} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{2}} {\overset{\propto} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by Olaf last updated on 16/Sep/20

Erratum! S_n =(1/2)( Σ_(k=2) ^n (1/(k−1))−Σ_(k=2) ^n (1/(k+1))) S_n =(1/2)( Σ_(k=1) ^(n−1) (1/k)−Σ_(k=3) ^(n+1) (1/k)) S_n =(1/2)( 1+(1/2)−(1/n)−(1/(n+1))) S_∞ = lim_(n→∞) S_n = (3/4)

$$\mathrm{Erratum}! \\ $$$$\mathrm{S}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}−\mathrm{1}}−\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$\mathrm{S}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}−\underset{{k}=\mathrm{3}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}\right) \\ $$$$\mathrm{S}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$\mathrm{S}_{\infty} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}S}_{{n}} \:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by mr W last updated on 16/Sep/20

you are right!

$${you}\:{are}\:{right}! \\ $$

Answered by Olaf last updated on 16/Sep/20

for n=1, (1/(n^2 −1)) is not defined!

$$\mathrm{for}\:{n}=\mathrm{1},\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}! \\ $$

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