Integral Calculus


Let f : R -> (0,∞) be a decreasing function. For n ∈ N.

We define,     a(n) = f(1) + f(2) + ... +f(n) - 1 ∫ n f .
Show that the infinite sequence (a(n)) is a bounded, monotonic one, and hence it converges.

4 comments:

  1. i somehow managed to prove the opposite :/.plz tell my mistake


    a(n) = f(1) + f(2) + ... +f(n) - 1 ∫ n f(x)dx

    differentiating wrt to n
    a'(n)= -f(n)

    again differentiating wrt to n
    a"(n)= -f'(n)

    f'(n)<0
    ⇒ a"(n)>0 (d^2a/dn^2 >0,concavity)
    ⇒ a(n)is concave for all n
    if it always remains concave,its graph will diverge.

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  2. write x^2 as x*x=x+x+x+x+x+x....(x times)

    differentiating wrt x..
    2x=1+1+1+1+1+1+1+1....(x times)
    =>2x=x

    absurd, isn't it?
    if you get your mistake from this, do tell.

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  3. where i see the mistake is in this thing-

    2x=1+1+1+1+1+1+1+1....(x times)

    if we put x a non intger.say 1.5,LHS=3 but we can't add 1,1.5 times.so i think the above equation is true ONLY for integers?

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