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

Question Number 2589 by Filup last updated on 23/Nov/15

f_{n} = \frac{1}{n} (n + f_{n - 1})

f_{1} = 1

Evaluate :

S = \underset{i = 1}{\sum^{m}} 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

: (o h w e l l

Commented by Filup last updated on 23/Nov/15

Oh . It seems I was over thinking .

Answered by 123456 last updated on 23/Nov/15

f_{n} = \frac{1}{n} (n + f_{n - 1}) = 1 + \frac{f_{n - 1}}{n}, n \neq 0

f_{n - 1} = n (f_{n} - 1)

f_{0} = 1 (f_{1} - 1) = 0

f_{1} = 1 + \frac{f_{0}}{1} = 1

f_{2} = 1 + \frac{f_{1}}{2} = 1 + \frac{1}{2} = \frac{3}{2}

f_{3} = 1 + \frac{f_{2}}{3} = 1 + \frac{1}{2} = \frac{3}{2} = 1 + \frac{1}{3} + \frac{1}{6}

f_{4} = 1 + \frac{f_{3}}{4} = 1 + \frac{3}{8} = \frac{11}{8} = 1 + \frac{1}{4} + \frac{1}{12} + \frac{1}{4!}

f_{5} = 1 + \frac{f_{4}}{5} = 1 + \frac{11}{40} = \frac{51}{40} = 1 + \frac{1}{5} + \frac{1}{20} + \frac{1}{60} + \frac{1}{5!}

- - - - continue - - - - -

Commented by 123456 last updated on 23/Nov/15

f_{2} = 1 + \frac{f_{1}}{2}

f_{3} = 1 + \frac{f_{2}}{3} = 1 + \frac{1}{3} + \frac{f_{1}}{6}

f_{4} = 1 + \frac{f_{3}}{4} = 1 + \frac{1}{4} + \frac{1}{12} + \frac{f_{1}}{4!}

f_{5} = 1 + \frac{f_{4}}{5} = 1 + \frac{1}{5} + \frac{1}{20} + \frac{1}{60} + \frac{f_{1}}{5!}

⋮

f_{n} = S_{n} + \frac{f_{1}}{n!} (n \in N^{*})

S_{n} = 1 + \frac{S_{n - 1}}{n}, S_{1} = 0, S_{2} = 1, \dots

Answered by prakash jain last updated on 23/Nov/15

f_{n} = 1 + \frac{1}{n} + \frac{1}{n (n - 1)} + \dots + \frac{1}{n!}

f_{n} = \frac{1}{n!} (n! + (n - 1)! + (n - 2)! + . . + 1)

f_{i} = \underset{k = 1}{\sum^{i}} \frac{k!}{i!}

S = \underset{i = 1}{\sum^{n}} \underset{k = 1}{\sum^{i}} \frac{k!}{i!}

I think a closed form expression for sum

may be possible in terms of gamma function .