Popescu and D. Schwartz Solution. Let F : # x [0, ∞) → [0,∞) be defined by F(x) =

C. Popescu and D. Schwartz Solution. Let F : # x [0, ∞) → [0,∞) be defined by F(x) =

0 f (t)dt. Since f takes on nonnegative values, it follows that F is increasing, so

lim F x →∞ (x) = sup {F(x) : x ≥ 0} > 0.

We now prove that lim x →∞ F (x) = ∞. Indeed, if this limit is finite and equals L > 0, then lim x →∞ f (x) = ℓ/L. Hence f (x) ≥ ℓ/(2L) for x ≥ x #

0 > 0. So, for all x ≥x 0 ,F (x) = F(x 0 )+ x x 0 f (t)dt ≥ ℓ(x − x 0 )/(2L) , which contradicts the assump-

tion that lim x →∞ F (x) is finite. Applying l’Hˆopital’s rule, we obtain

F 2 (x)

x lim = 2 lim f (x)F (x) = 2ℓ .

x →∞

346 9 Riemann Integrability It follows that lim

F (x)

x →∞ √ x = 2 ℓ . Finally, we write

f (x)F(x)

f (x) x =

F √ , (x)

√ for x large enough, to conclude that lim x →∞ f (x) x exists, is finite, and equals

9.5.9. Assume that f is a real-valued continuously differentiable function on some interval [a, ∞) such that f ′ (x) + α f (x) tends to zero as x tends to infinity, for some α >0 . Prove that f (x) tends to zero as x tends to infinity. Solution. Set g : =f ′ + α f . Arguing by contradiction, there exist C 1 > 0 and a sequence (x n ) n ≥1 such that lim n → ∞ x n = ∞ and | f (x n )| ≥ C 1 , for all n ≥ 1. Without loss of generality, we can assume that f (x n )≥C 1 , for all n ≥ 1. Hence f ′ (x n )=

g (x n )− α f (x n ) ≤ g(x n )− α C 1 . So, since g (x)→0 as x→∞ and α > 0, there exists N 0 ∈ N such that f ′ (x n )≤− α C 1 /2 ≡ −C 2 , for all n ≥N 0 . Therefore

f ′ (x n )

f (x ) ≤− n C ≡ −C 3 1 ,

for all n ≥N 0 , where C 3 > 0. By integration on [a, x n ] in the above inequality we obtain

ln f (x n ) ≤ −C 3 x n +C 4 ,

or equivalently,

f (x n )≤e −C 3 x n +C 4 ,

for all n ≥N 0 . This implies f (x n )→0 as n→∞, which contradicts our assumption

f (x n )≥C 1 > 0, for all n ≥ 1. In conclusion, lim x → ∞ f (x) = 0. It is obvious that the statement does not remain true if α ≤ 0. Give an example! ⊓ ⊔

9.5.10. (i) Prove that for any positive integer n ,

(n − 1)! ≤ n n e −n e ≤ n!.

(ii) Deduce that

1 (n!) /n

1 /n

approaches e −1 as n tends to infinity. Solution. (i) A direct proof uses an induction argument based on the inequalities

1 +1 1 n

<e< 1 +

An alternative proof using the integral calculus uses the evaluation and compari- son of # n

1 ln xdx with the upper and lower Darboux sums associated with the partition (1, 2, . . . , n) of the interval [1, n].

9.5 Riemann Iintegrals and Limits 347 (ii) Taking the nth root of the right inequality in (i), we obtain

n e −1 e 1 /n

1 ≤ (n!) /n .

Dividing by n yields

(n!) 1 /n

e 1 /n

On the other hand, we multiply the first inequality in (i) by n and take the nth root. Therefore

(n!) 1 /n ≤ ne −1 e 1 /n n 1 /n .

Dividing by n yields

(n!) 1 /n

≤ e 1 /n n 1 /n .

But both n 1 /n and e 1 /n approach 1 as n becomes large. Thus our quotient is squeezed between two numbers approaching e −1 , and must therefore approach e −1 . ⊓ ⊔

9.5.11. Let g : [0, 1] → R be a continuous function such that lim x →0+ g (x)/x exists and is finite. Prove that

D. Andrica and M. Piticari Solution. Define the function h : [0, 1] → R by

g (x) x →0,x>0 x

if t = 0.

Then h is continuous, and we can set

H (x) =

h (t)dt .

We have 0

g (x n )dx = n

h 1 (x n )dx = xH(x n ) −

H (x n )dx

dx = H(1) − n

" 1 n " 1 g (x)

H (x )dx =

H (x )dx.

If 0 < a < 1, then " 1 " 1 " a " 1

H (x n )dx ≤

|H(x n )|dx =

|H(x n )|dx +

|H(x n )|dx

0 0 0 a (9.12) ≤ a|H( α n n )| + (1 − a)M,

where α n ∈ [0,a] and M = max t ∈[0,1] |H(t)|.

348 9 Riemann Integrability

Consider ε > 0 such that a > 1 − ε /(2M). Since lim n →∞ |H( α n

n )| = 0, it follows that a |H( α n n )| < 2 for all positive integers n ≥ N( ε ). Relation (9.12) yields

Hence lim # 1

n →∞ 0 H (x n )dx = 0 and the conclusion follows. ⊓ ⊔