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f-n-1-n-n-f-n-1-f-1-1-Evaluate-S-i-1-m-f-i-

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Question Number 2589 by Filup last updated on 23/Nov/15
f_n =(1/n)(n+f_(n−1) ) f_1 =1 Evaluate: S=Σ_(i=1) ^m f_i
$${f}_{{n}} =\frac{\mathrm{1}}{{n}}\left({n}+{f}_{{n}−\mathrm{1}} \right) \\ $$$${f}_{\mathrm{1}} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Evaluate}: \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}{f}_{{i}} \\ $$
Commented by 123456 last updated on 23/Nov/15
?????
$$????? \\ $$
Commented by Filup last updated on 23/Nov/15
my working was incorrect i belive :( oh well
$$\mathrm{my}\:\mathrm{working}\:\mathrm{was}\:\mathrm{incorrect}\:\mathrm{i}\:\mathrm{belive} \\ $$$$:\left(\:{oh}\:{well}\right. \\ $$
Commented by Filup last updated on 23/Nov/15
Oh. It seems I was over thinking.
$$\mathrm{Oh}.\:\mathrm{It}\:\mathrm{seems}\:\mathrm{I}\:\mathrm{was}\:\mathrm{over}\:\mathrm{thinking}.\: \\ $$
Answered by 123456 last updated on 23/Nov/15
f_n =(1/n)(n+f_(n−1) )=1+(f_(n−1) /n),n≠0 f_(n−1) =n(f_n −1) f_0 =1(f_1 −1)=0 f_1 =1+(f_0 /1)=1 f_2 =1+(f_1 /2)=1+(1/2)=(3/2) f_3 =1+(f_2 /3)=1+(1/2)=(3/2)=1+(1/3)+(1/6) f_4 =1+(f_3 /4)=1+(3/8)=((11)/8)=1+(1/4)+(1/(12))+(1/(4!)) f_5 =1+(f_4 /5)=1+((11)/(40))=((51)/(40))=1+(1/5)+(1/(20))+(1/(60))+(1/(5!)) −−−−continue−−−−−
$${f}_{{n}} =\frac{\mathrm{1}}{{n}}\left({n}+{f}_{{n}−\mathrm{1}} \right)=\mathrm{1}+\frac{{f}_{{n}−\mathrm{1}} }{{n}},{n}\neq\mathrm{0} \\ $$$${f}_{{n}−\mathrm{1}} ={n}\left({f}_{{n}} −\mathrm{1}\right) \\ $$$${f}_{\mathrm{0}} =\mathrm{1}\left({f}_{\mathrm{1}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${f}_{\mathrm{1}} =\mathrm{1}+\frac{{f}_{\mathrm{0}} }{\mathrm{1}}=\mathrm{1} \\ $$$${f}_{\mathrm{2}} =\mathrm{1}+\frac{{f}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}_{\mathrm{3}} =\mathrm{1}+\frac{{f}_{\mathrm{2}} }{\mathrm{3}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${f}_{\mathrm{4}} =\mathrm{1}+\frac{{f}_{\mathrm{3}} }{\mathrm{4}}=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{11}}{\mathrm{8}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}!} \\ $$$${f}_{\mathrm{5}} =\mathrm{1}+\frac{{f}_{\mathrm{4}} }{\mathrm{5}}=\mathrm{1}+\frac{\mathrm{11}}{\mathrm{40}}=\frac{\mathrm{51}}{\mathrm{40}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{60}}+\frac{\mathrm{1}}{\mathrm{5}!} \\ $$$$−−−−\mathrm{continue}−−−−− \\ $$
Commented by 123456 last updated on 23/Nov/15
f_2 =1+(f_1 /2) f_3 =1+(f_2 /3)=1+(1/3)+(f_1 /6) f_4 =1+(f_3 /4)=1+(1/4)+(1/(12))+(f_1 /(4!)) f_5 =1+(f_4 /5)=1+(1/5)+(1/(20))+(1/(60))+(f_1 /(5!)) ⋮ f_n =S_n +(f_1 /(n!)) (n∈N^∗ ) S_n =1+(S_(n−1) /n),S_1 =0,S_2 =1,...
$${f}_{\mathrm{2}} =\mathrm{1}+\frac{{f}_{\mathrm{1}} }{\mathrm{2}} \\ $$$${f}_{\mathrm{3}} =\mathrm{1}+\frac{{f}_{\mathrm{2}} }{\mathrm{3}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{{f}_{\mathrm{1}} }{\mathrm{6}} \\ $$$${f}_{\mathrm{4}} =\mathrm{1}+\frac{{f}_{\mathrm{3}} }{\mathrm{4}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{{f}_{\mathrm{1}} }{\mathrm{4}!} \\ $$$${f}_{\mathrm{5}} =\mathrm{1}+\frac{{f}_{\mathrm{4}} }{\mathrm{5}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{60}}+\frac{{f}_{\mathrm{1}} }{\mathrm{5}!} \\ $$$$\vdots \\ $$$${f}_{{n}} =\mathrm{S}_{{n}} +\frac{{f}_{\mathrm{1}} }{{n}!}\:\:\:\:\:\left({n}\in\mathbb{N}^{\ast} \right) \\ $$$$\mathrm{S}_{{n}} =\mathrm{1}+\frac{\mathrm{S}_{{n}−\mathrm{1}} }{{n}},\mathrm{S}_{\mathrm{1}} =\mathrm{0},\mathrm{S}_{\mathrm{2}} =\mathrm{1},… \\ $$
Answered by prakash jain last updated on 23/Nov/15
f_n =1+(1/n)+(1/(n(n−1)))+...+(1/(n!)) f_n =(1/(n!))(n!+(n−1)!+(n−2)!+..+1) f_i =Σ_(k=1) ^i ((k!)/(i!)) S=Σ_(i=1) ^n Σ_(k=1) ^i ((k!)/(i!)) I think a closed form expression for sum may be possible in terms of gamma function.
$${f}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}\left({n}−\mathrm{1}\right)}+…+\frac{\mathrm{1}}{{n}!} \\ $$$${f}_{{n}} =\frac{\mathrm{1}}{{n}!}\left({n}!+\left({n}−\mathrm{1}\right)!+\left({n}−\mathrm{2}\right)!+..+\mathrm{1}\right) \\ $$$${f}_{{i}} =\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{{k}!}{{i}!} \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{{k}!}{{i}!} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{sum} \\ $$$$\mathrm{may}\:\mathrm{be}\:\mathrm{possible}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{gamma}\:\mathrm{function}. \\ $$

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