MathDB

1

Part of 1988 Iran MO (2nd round)

Problems(2)

The sum of terms is less than 1 - [Iran Second Round 1988]

Source:

12/7/2010
Let {an}n=1∞\{a_n \}_{n=1}^{\infty} be a sequence such that a1=12a_1=\frac 12 and an=(2n−32n)an−1∀n≥2.a_n=\biggl( \frac{2n-3}{2n} \biggr) a_{n-1} \qquad \forall n \geq 2. Prove that for every positive integer n,n, we have ∑k=1nak<1.\sum_{k=1}^n a_k <1.
number theory proposednumber theory
The fraction has a limit - [Iran Second Round 1988]

Source:

12/7/2010
(a) Prove that for all positive integers m,nm,n we have ∑k=1nk(k+1)(k+2)⋯(k+m−1)=n(n+1)(n+2)⋯(n+m)m+1\sum_{k=1}^n k(k+1)(k+2)\cdots (k+m-1)=\frac{n(n+1)(n+2) \cdots (n+m)}{m+1}
(b) Let P(x)P(x) be a polynomial with rational coefficients and degree m.m. If nn tends to infinity, then prove that ∑k=1nP(k)nm+1\frac{\sum_{k=1}^n P(k)}{n^{m+1}} Has a limit.
algebrapolynomialinductionfactorialalgebra proposed