About Factorial Sums

About Factorial Sums Mihály Bencze1 and Florentin Smarandache2 Str. Hărmanului 6, 505600 Săcele-Négyfalu, Jud. Braşov, Romania 2 Chair of Math & sciences, University of New-Mexico, 200 College Road, NM 87301, USA 1

Abstract. In this paper, we present some new inequalities for factorial sum. Application 1.We have the following inequality n 2((n + 1)!− 1) k!≤ ∑ n +1 k =1 Proof. If xk , yk > 0 (k = 1, 2,..., n) , have the same monotonity, then ⎛1 n ⎞⎛ 1 n ⎞ 1 n x y ⎜ n ∑ k ⎟⎜ n ∑ k ⎟ ≤ n ∑ xk yk k =1 ⎝ k =1 ⎠⎝ k =1 ⎠

(1)

the Chebishev’s inequality. If xk , yk have different monotonity, then holds true the reverse inequality, we take xk = k , yk = k ! (k = 1, 2,..., n) and use that

n

∑k ⋅k! =

(n + 1)! − 1 .

k =1

Application 2. We have the following inequality n 3(n + 1)(n + 1)! k!≤ ∑ n 2 + 3n + 5 k =1 Proof. In (1) we take xk = k 2 + k + 1 ; yk = k ! (k = 1, 2,..., n) and the identity n

∑ (k

2

+ k + 1)k ! = (n + 1)(n + 1)!

k =1

Application 3. We have the following inequality n 1 n 2 (n + 1) ≥ ∑ 2((n + 1)!− 1) k =1 k ! Proof. Using the Application 1, we take 1

n

1

∑ k! ≥ k =1

n2 n

∑k!

≥

n 2 (n + 1) 2((n + 1)!− 1)

k =1

Application 4. We have the following inequality n 1 n 2 (n 2 + 3n + 5) ≥ ∑ 3(n + 1)(n + 1)! k =1 k ! Proof. Using the Application 2, we take n 1 n2 n 2 (n 2 + 3n + 5) ≥ n ≥ ∑ 3(n + 1)(n + 1)! k =1 k ! ∑k! k =1

Application 5. We have the following inequality: n 1 2⎛ 1⎞ ≥ 1 + ⎜1 − ⎟ ∑ n ⎝ n! ⎠ k =1 k ! 1 Proof. In (1) we take xk = k , yk = , (k = 1, 2,..., n) and we obtain (k + 1)! 1 ⎛ n ⎞⎛ n 1 ⎞ n k 1 k = 1− ∑ ⎜∑ ⎟≥∑ ⎜ ⎟ n ⎝ k =1 ⎠ ⎝ k =1 (k + 1)! ⎠ k =1 (k + 1)! (n + 1)! therefore ⎛ n 1 ⎞ 2 ⎛ 1 ⎞ ⎜∑ ⎟≥ ⎜1 − ⎟ ⎝ k =1 (k + 1)! ⎠ n + 1 ⎝ (n + 1)! ⎠ or n 1 2⎛ 1⎞ ≥ ⎜1 − ⎟ ∑ n ⎝ n! ⎠ k =2 k ! therefore 2⎛ 1⎞ ⎛ n 1⎞ ⎜ ∑ k ! ⎟ ≥ 1 + n ⎜1 − n ! ⎟ ⎝ ⎠ ⎝ k =1 ⎠ Application 6. We have the following inequality: ⎛ n ⎞ 1 2 ⎛ 1 ⎞ ⎜∑ ⎟≥ ⎜1 − ⎟ 2 ⎝ k =1 (k + 2) k ! ⎠ n + 5 ⎝ (n + 2)! ⎠ 1 Proof. In (1) we take xk = k + 2, yk = , (k = 1, 2,..., n) (k + 2) 2 k ! therefore n 1⎛ n 1 1 1 ⎞ n + ≥ = 1− k ( 2) ∑ ∑ ∑ ⎜ ⎟ 2 2 n ⎝ k =1 (n + 2)! ⎠ k =1 (k + 2) k ! k =1 (k + 2) k ! therefore 2

n

2 ⎛

1

1

⎞

∑ (k + 2) k ! ≥ n + 5 ⎜1 − (n + 2)! ⎟ k =1

2

⎝

⎠

Application 7.We have the following inequality: n ⎛1 ⎞ 1 6 1 ≥ 2 ∑ ⎜ − ⎟ 2n + 9n + 1 ⎝ 2 (n + 1)(n + 2)! ⎠ k =1 k ( k + 1)( k + 2)! Proof. In (1) we take 1 xk = k 2 + 2k + 2, yk = , (k = 1, 2,..., n) k (k + 1)(k + 2)!

then n n 1 n 2 1 k 2 + 2k + 2 ( k + 2 k + 2) ≥ = ∑ ∑ ∑ n k =1 k =1 k ( k + 1)( k + 2)! k =1 k ( k + 1)( k + 2)! n 1 1 1 1 =∑ − = − (k + 1)(k + 2)! 2 (n + 1)(n + 2)! k =1 k ( k + 1)!

Application 8. We have the following inequality: n 1 n ≥ 2 ∑ 4 2n + 2n + 1 k =1 4k + 1 1 Proof. In (1) we take xk = 4k , yk = 4 , (k = 1, 2,..., n) , 4k + 1 therefore n 1⎛ n 1 ⎞ n 4k 1 1 2n(n + 1) ⎞⎛ n ⎛ ⎞ 4k ⎟ ⎜ ∑ 4 ⎟ ≥ ∑ 4 = ∑⎜ 2 − 2 ∑ ⎟= 2 ⎜ n ⎝ k =1 ⎠ ⎝ k =1 4k + 1 ⎠ k =1 4k + 1 k =1 ⎝ 2k − 2k + 1 2k + 2k + 1 ⎠ 2n + 2n + 1 Application 9. We have the following inequality: n 1 3n ≥ ∑ 4 (2n + 1) 2 k =1 4k − 1 1 Proof. In (1) we take xk = k 2 , yk = 2 , (k = 1, 2,..., n) then 4k − 1 n n 1⎛ 1 ⎞ n k2 n(n + 1) 2 ⎞⎛ ≥∑ 2 = , etc. k ∑ ∑ ⎜ ⎟ ⎜ ⎟ 2 n ⎝ k =1 ⎠ ⎝ k =1 4k − 1 ⎠ k =1 4k − 1 2(2n + 1) Reference:

[1] Octogon Mathematical Magazine (1993-2007) {Published in Octogon Mathematical Magazine, Vol. 15, No. 2, 810-812, 2007.} 3