Stolz Cesaro Lý thuyết và một số ứng dụng

Lý thuyết về định lý Stolz Cesaro. Phát biểu của định lý Stolz Cesaro. Cách chứng minh định lý Stolz Cesaro. Một số kết quả ứng dụng được. Một số dãy số đặc biệt. Một số kết quả quen thuộc. Định lý LHopitale. Tài liệu được viết bằng tiếng Anh Nguồn từ đại học Kansas.

The purpose of this note is to show how obtain limit or equivalent of sequence with Theorem Stolz-Cesaro by Moubinool OMARJEE Let me remind the main tool in this note _ Theorem of Stolz-Cesaro: a n  n≥0 , b n  n≥0 sequences of reals numbers Assume b n  n≥0 is strictly increasing , unbounded and the following limit exist lim a n+1 − a n = L n→+∞ b n+1 − b n Then the lim a n also exist and its equal to L bn n→+∞ APPLICATION : x n  n≥1 sequence of real number with x n > for any n ≥ lim x n = +∞ n→+∞ n Then n ∑ lim n→+∞ n k=1 =0 xn Proof: With S.C (Stolz-Cesaro) n lim n→+∞ n ∑ k=1 = lim n→+∞ xn x n+1 = n→+∞ lim n+1 − n n+1 + n n+1 = n→+∞ lim =0 x n+1 x n+1 APPLICATION : x n  n≥1 sequence of real number given by xn = n k k! ∑ k=1 n + 1! − Find the limit of x n  n≥1 Proof: With S.C lim x n→+∞ n = n→+∞ lim n + 1n + 1! n + 1n + 1! = n→+∞ lim =1 n + 2! − n + 1! n + 1n + 1! APPLICATION :x n  n≥1 sequence of real number with x = x n+1 = x + x + +x n Find lim x n n→+∞ n Proof: All the term of x n  n≥1 are strictly positive x 2n+1 = x 2n + x n x n  n≥1 is strictly increasing, if it was bounded then lim x n→+∞ n exist = L L2 = L2 + L This gives the limit L = contradiction since all x n ≥ So lim x n→+∞ n = +∞ x 2n+1 = + x1n → x 2n lim x n+1 = n→+∞ x n With S.C theorem x n = lim x n+1 − x n = lim x 2n+1 − x 2n lim n→+∞ n n→+∞ n + 1 − n n→+∞ x n+1 + x n = n→+∞ lim x x n+ x = n→+∞ lim x n+11 = = n+1 n 1+1 + xn APPLICATION :Find lim n→+∞ lnn! − n lnn n Proof: With S.C lnn! − n lnn lnn + 1 − n + 1 lnn + 1 + n lnn lim = n→+∞ lim n n→+∞ n + 1 − n = n→+∞ lim −n lnn + 1 + n lnn = n→+∞ lim − ln1 + n n  = −1 APPLICATION 5: Find n Ene n  ∑ k=1 lim n2 n→+∞ Proof: With S.C n Eke k  ∑ k=1 lim n→+∞ n2 Ene n  n = n→+∞ lim = n→+∞ lim = 2n + n − n − 1 APPLICATION 6: x n  n≥1 sequence of real number with n lim x x2 = n→+∞ n ∑ k k=1 Prove that lim 3n x n = n→+∞ Proof: S n = ∑ k=1 x 2k , x n S n  tend to The sequence S n  is strictly increasing n if S n  converge to L , then x n → L1 > and we will have S n = ∑ k=1 x 2k → +∞ contradiction We have n lim S n = +∞ and n→+∞ lim x n = n→+∞ We also have x n S n−1 → when n → +∞ and S 3n − S 3n−1 = x 2n S 2n + S n S n−1 + S 2n−1  → when n → +∞ With S.C S − S 3n−1 S3 lim n =3 lim nn = n→+∞ n − n − 1 n→+∞ We deduce that 3 3n →1 Sn 3n x n = 3n x S →1 Sn n n APPLICATION 7: x n  n≥1 sequence of reals numbers with < x < and x n+1 = x n − x 2n for any n ≥ Prove that lim nx n = n→+∞ Proof: The sequence x n  n≥1 is decreasing and bounded lim x n→+∞ n = L exist L = L − L2 So L = Consider y n = xn with S.C y y − yn x n+1 lim n = n→+∞ lim n+1 = n→+∞ lim x nx n−x n+1 n→+∞ n n + 1 − n x 2n = n→+∞ lim = n→+∞ lim =1 − xn x n x n − x 2n  APPLICATION 8: x n  n≥1 sequence of reals numbers with x = x n+1 = x 2n +2 n+1 for any n ≥ Find n xk ∑ k=1 lim n→+∞ ∑ n k=1 k Proof: By induction < x n ≤ for any n ≥ 2 < x n+1 ≤ + → when n → +∞ n+1 With S.C n xk ∑ k=1 x n = lim n x 2n−1 + = lim = lim n n→+∞ ∑ n n→+∞ n→+∞ n k=1 k APPLICATION 9: x n  n≥1 sequence of reals numbers such that lim x n+1n− x n = L > n→+∞ Prove that lim n→+∞ n+1 x n+1 − xn n n! n + 1! = eL Proof: Since x n+1 − x n ∼ nL the equivalence of partial sum for divergence n−1 n−1 gives ∑ k=1 x k+1 − x k  ∼ L ∑ k=1 k this gives xnn ∼ nL With S.C x n+1 − nx n n+1 n+1! n! x n+1 lim − xn = n→+∞ lim n→+∞ n n! n + 1 − n n+1 n + 1! = n→+∞ lim x n = eL n n n! Since n n! = e n − e lnn+ln2π n2 +O n3