1.f(n)=1n[(n+1)(n+2)⋯(2n)]1/nf(n)=\frac1n[(n+1)(n+2)\cdots(2n)]^{1/n}f(n)=n1[(n+1)(n+2)⋯(2n)]1/n, limn→∞f(n)=\lim_{n\to\infty} f(n) =limn→∞f(n)=a.4/e4/e4/eb.log(4/e)\log(4/e)log(4/e)c.e/4e/4e/4d.log2\log2log2Login to continueOnly logged in users canattempt or see the solution.
