Limit of a sequence

Calculus Level 5

Let us define k0=0k_0=0 and kn+1=kn+1+kn2k_{n+1}=k_n+\sqrt{1+k_n^2}.

Find the value of limn(kn2n)\lim_{n \rightarrow \infty} \left(\dfrac{k_n}{2^n}\right)

Also try Limit of a sequence (2) and (3)

Assumptions and Source\textbf{Assumptions and Source}

  • nn is any non-negative integer.

  • This is a standard problem in most calculus textbooks for JEE (Advanced)\text{JEE (Advanced)}.


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