Question

数列求解

Answer 1

f(x)=x+1/x (1)
f'(x) =1-1/x^2
f'(n) =n-1/n^2
ln:equation of tangent at (n, f(n))
y-f(n)= f'(n) (x-n)
y- (n+1/n) = (1-1/n^2) (x-n) (2)

x=n+1 (3)
sub (3) into (1)
y=(n+1) + 1/(n+1)

ie
An =(n+1, (n+1) + 1/(n+1))

sub (3) into (2)
y- (n+1/n) = (1-1/n^2) (n+1-n)
y = (1-1/n^2) +(n+1/n)
= (n+1) + (n-1)/n^2
ie
Bn=(n+1, (n+1) + (n-1)/n^2)
an =|AnBn|
=|(n-1)/n^2 -1/(n+1)|
=| -1/[n^2.(n+1)]|
=1/[n^2.(n+1)]
cn =nan = 1/[n(n+1)] = 1/n -1/(n+1)
Sn =c1+c2+...+cn
=1 - 1/(n+1)
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