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Sum-1-1-1-1-2-1-1-2-3-1-1-2-3-8016-




Question Number 79635 by TawaTawa last updated on 26/Jan/20
Sum:  (1/1) + (1/(1 + 2)) + (1/(1 + 2 + 3)) + ...  +  (1/(1 + 2 + 3 + ... + 8016))
$$\mathrm{Sum}:\:\:\frac{\mathrm{1}}{\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:…\:+\:\mathrm{8016}} \\ $$
Commented by mr W last updated on 26/Jan/20
=Σ_(k=1) ^n (1/(1+2+3+...+k))  =2Σ_(k=1) ^n (1/(k(k+1)))  =2Σ_(k=1) ^n ((1/k)−(1/(k+1)))  =2(1−(1/(n+1)))  =((2n)/(n+1))  =((2×8016)/(8017))  =((16032)/(8017))    if n→∞  Σ→2
$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}} \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{2}{n}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}×\mathrm{8016}}{\mathrm{8017}} \\ $$$$=\frac{\mathrm{16032}}{\mathrm{8017}} \\ $$$$ \\ $$$${if}\:{n}\rightarrow\infty \\ $$$$\Sigma\rightarrow\mathrm{2} \\ $$
Commented by TawaTawa last updated on 26/Jan/20
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 26/Jan/20
I did not understand:   2 Σ_(k = 1) ^n  (1/(k(k + 1)))  sir
$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{understand}:\:\:\:\mathrm{2}\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}\:+\:\mathrm{1}\right)}\:\:\mathrm{sir} \\ $$
Commented by mr W last updated on 26/Jan/20
you don′t know 1+2+3+...+k=?
$${you}\:{don}'{t}\:{know}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}=? \\ $$
Commented by mr W last updated on 26/Jan/20
Commented by TawaTawa last updated on 26/Jan/20
ohh. i understand now sir.      1 + 2 + 3 + .. + n  =  ((n(n + 1))/2)   ⇒ (1/((n(n + 1))/2))   =  (2/(n(n + 1)))   so you take the 2 out.
$$\mathrm{ohh}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}. \\ $$$$\:\:\:\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:..\:+\:\mathrm{n}\:\:=\:\:\frac{\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{1}\right)}{\mathrm{2}}\:\:\:\Rightarrow\:\frac{\mathrm{1}}{\frac{\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{1}\right)}{\mathrm{2}}}\:\:\:=\:\:\frac{\mathrm{2}}{\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{1}\right)}\: \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{take}\:\mathrm{the}\:\mathrm{2}\:\mathrm{out}. \\ $$
Commented by TawaTawa last updated on 26/Jan/20
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by john santu last updated on 27/Jan/20
1+(1/3)+(1/6)+(1/(10))+(1/(15))+(1/(21))+...=S_m   consider denumerator  1,3,6,10,15,21,...  T_n = ((n^2 +n)/2) = ((n(n+1))/2)  Σ_(n=1 ) ^(8016)  (2/(n(n+1))) = 2 Σ_(n=1) ^(8016)  (1/n)−(1/(n+1))  = 2 {(1/1)−(1/(8017))} = ((16032)/(8017))
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{21}}+…=\mathrm{S}_{\mathrm{m}} \\ $$$$\mathrm{consider}\:\mathrm{denumerator} \\ $$$$\mathrm{1},\mathrm{3},\mathrm{6},\mathrm{10},\mathrm{15},\mathrm{21},… \\ $$$$\mathrm{T}_{\mathrm{n}} =\:\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}{\mathrm{2}}\:=\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\underset{\mathrm{n}=\mathrm{1}\:} {\overset{\mathrm{8016}} {\sum}}\:\frac{\mathrm{2}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:=\:\mathrm{2}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{8016}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$=\:\mathrm{2}\:\left\{\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8017}}\right\}\:=\:\frac{\mathrm{16032}}{\mathrm{8017}} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 27/Jan/20
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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