Daniel: Allow me to explain that part. In the first problem, when evaluating the limit, we notice that the denominator is a polynomial in n. (k is just a constant.) And since the term n^k in the denominator dominates the other polynomial terms, it makes sense to divide both the denominator and the numerator by n^k to get a constant term (k+1, k) in the denominator. (k+1, k) is just "k+1 choose k" or number of ways of choosing k objects out of k+1 distinct objects. And then in the last step, when taking limits, the numerator just goes to 1 while the denominator stays k+1.
To prove Stolz-Cesàro, you can extend the sequences to piecewise-linear functions by linearly interpolatiing between n and n+1, and then notice that L’Hospital’s rule holds for continuous piecewise-differentiable functions.
I didn't understand the last equality before taking the limit in the first problem :-|. Anyway, it is a nice result.
ReplyDeleteDaniel: Allow me to explain that part. In the first problem, when evaluating the limit, we notice that the denominator is a polynomial in n. (k is just a constant.) And since the term n^k in the denominator dominates the other polynomial terms, it makes sense to divide both the denominator and the numerator by n^k to get a constant term (k+1, k) in the denominator. (k+1, k) is just "k+1 choose k" or number of ways of choosing k objects out of k+1 distinct objects. And then in the last step, when taking limits, the numerator just goes to 1 while the denominator stays k+1.
ReplyDeleteHope this helps. Thanks!
To prove Stolz-Cesàro, you can extend the sequences to piecewise-linear functions by linearly interpolatiing between n and n+1, and then notice that L’Hospital’s rule holds for continuous piecewise-differentiable functions.
ReplyDeleteThanks for the explanation, Vishal!
ReplyDelete