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If-r-1-n-t-r-n-n-1-n-2-n-3-8-then-lim-n-r-1-n-1-t-r-




Question Number 110173 by ajfour last updated on 27/Aug/20
If  Σ_(r=1) ^n t_r =((n(n+1)(n+2)(n+3))/8)  then    lim_(n→∞)   Σ_(r=1) ^n  (1/t_r ) = ?
$${If}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{t}_{{r}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{8}} \\ $$$${then}\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{t}_{{r}} }\:=\:? \\ $$$$ \\ $$
Answered by Olaf last updated on 27/Aug/20
t_n  = Σ_(r=1) ^n t_r −Σ_(r=1) ^(n−1) t_r   t_n  = ((n(n+1)(n+2)(n+3))/8)−(((n−1)n(n+1)(n+2))/8)  t_n  = ((n(n+1)(n+2)[n+3−n+1])/8)  t_n  = ((n(n+1)(n+2))/2)  t_r  = ((r(r+1)(r+2))/2)  (1/t_r ) = (2/(r(r+1)(r+2))) = (1/r)−(2/(r+1))+(1/(r+2))  Σ_(r=1) ^n (1/t_r ) = Σ_(r=1) ^n (1/r)−2Σ_(r=1) ^n (1/(r+1))+Σ_(r=1) ^n (1/(r+2))  Σ_(r=1) ^n (1/t_r ) = Σ_(r=1) ^n (1/r)−2Σ_(r=2) ^(n+1) (1/r)+Σ_(r=3) ^(n+2) (1/r)  Σ_(r=1) ^n (1/t_r ) = Σ_(r=1) ^n (1/r)−2(Σ_(r=1) ^n (1/r)−1+(1/(n+1)))  +(Σ_(r=1) ^n (1/r)−1−(1/2)+(1/(n+1))+(1/(n+2)))  Σ_(r=1) ^n (1/t_r ) = 2−(2/(n+1))−(3/2)+(1/(n+1))+(1/(n+2))  Σ_(r=1) ^n (1/t_r ) = (1/2)−(1/(n+1))+(1/(n+2))  Σ_(r=1) ^n (1/t_r ) = (1/2)−(1/((n+1)(n+2)))  lim_(n→∞) Σ_(r=1) ^n (1/t_r ) = (1/2)
$${t}_{{n}} \:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{t}_{{r}} −\underset{{r}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{t}_{{r}} \\ $$$${t}_{{n}} \:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{8}}−\frac{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{8}} \\ $$$${t}_{{n}} \:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left[{n}+\mathrm{3}−{n}+\mathrm{1}\right]}{\mathrm{8}} \\ $$$${t}_{{n}} \:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$${t}_{{r}} \:=\:\frac{{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\frac{\mathrm{2}}{{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{{r}}−\frac{\mathrm{2}}{{r}+\mathrm{1}}+\frac{\mathrm{1}}{{r}+\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}}−\mathrm{2}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}+\mathrm{1}}+\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}+\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}}−\mathrm{2}\underset{{r}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{r}}+\underset{{r}=\mathrm{3}} {\overset{{n}+\mathrm{2}} {\sum}}\frac{\mathrm{1}}{{r}} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}}−\mathrm{2}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}}−\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$+\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\mathrm{2}−\frac{\mathrm{2}}{{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{t}_{{r}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by ajfour last updated on 27/Aug/20
right answer, thanks Sir.
$${right}\:{answer},\:{thanks}\:{Sir}. \\ $$
Answered by Dwaipayan Shikari last updated on 27/Aug/20
△Σ_(r=1) ^n t_r =((n(n+1)(n+2)(n+3))/8)−(((n−1)n(n+1)(n+2))/8)  t_n =((n(n+1)(n+2))/2)  Σ^∞ (1/t_n )=Σ^∞ (2/(n(n+1)(n+2)))=Σ^∞ ((n+2−n)/(n(n+1)(n+2)))=Σ^∞ (1/n)−(1/(n+1))−Σ^∞ (1/(n+1))−(1/(n+2))  =lim_(n→∞) (1−(1/(n+1)))−lim_(n→∞) ((1/2)−(1/(n+2)))  =1−(1/2)=(1/2)
$$\bigtriangleup\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{t}_{{r}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{8}}−\frac{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{8}} \\ $$$${t}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\overset{\infty} {\sum}\frac{\mathrm{1}}{{t}_{{n}} }=\overset{\infty} {\sum}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\overset{\infty} {\sum}\frac{{n}+\mathrm{2}−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 27/Aug/20
Great way Sir; thanks a lot.
$${Great}\:{way}\:{Sir};\:{thanks}\:{a}\:{lot}. \\ $$

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