sophisticated limits with focus on mathematical olympiad problems By tetyana darian A thesis submitted to the

Graduate School-Camden

Rutgers, The State University of New Jersey

In partial fulfillment of the requirements

For the degree of Master of Science

Graduate Program in Pure Mathematics

Written under the direction of

Gabor Toth

And approved by

Gabor Toth

Camden, New Jersey

October 2019 thesis abstract Sophisticated Limits with Focus on Mathematical Olympiad Problems by tetyana darian Thesis Director:

Gabor Toth

The purpose of this thesis is to derive a variety of sophisticated limits, some in

Mathematical Olympiad Problems, without using mathematically advanced concepts.

In fact, this thesis requires only basic calculus and the Stolz-Ces`arotheorems. The

latter will be presented without proofs but with all the necessary ingredients and

formulas needed in this work. The thesis is written for a mathematically mature

student with a certain level of preparation to tackle these beautiful limits.

ii Acknowledgements

I would like to thank my thesis advisor, Professor Gabor Toth, for his thoughtful

guidance, support, and intuition. I truly appreciate and have enjoyed all of our

meetings, discussions, and classroom time together. Thank you for introducing me to

the beautiful and complex world of mathematics and mathematical research.

A separate thank you and appreciation goes to my husband, Steven, who supported

me through thick and thin during my mathematical journey.

iii Contents

Thesis Abstract ii

Acknowledgements iii

1 Introduction 1

1.1 Completeness of the Real Numbers ...... 1

1.2 Neighborhoods and Limit Points ...... 2

1.3 The Limit of a Sequence ...... 2

2 Stolz-Ces`aroTheorems and the Stirling Formula 4

2.1 The Stolz-Ces`aroTheorem ...... 4

2.2 The Stolz-Ces`aroTheorem ...... 5

2.3 The Stirling Formula ...... 7

3 Examples of Sophisticated Limits 8

3.1 Preliminaries ...... 8

3.2 Example 1 ...... 9

3.3 Example 2 ...... 13

3.4 Example 3 ...... 17

3.5 Example 4 ...... 20

3.6 Example 5 ...... 21

3.7 Example 6 ...... 23

Bibliography 25

iv 1

Chapter 1

Introduction

This chapter is a brief look at what this thesis is all about. To aid us in our discussion, we will be including some definitions and results which will be stated without proof.

Important: All definitions, propositions, and theorems presented in this chapter will be reintroduced, and elaborated upon throughout the course of the thesis. The

results of this chapter will be used without any explicit references. To further

emphasize this, we will not number the definitions, propositions, and theorems of

this chapter.

1.1 Completeness of the Real Numbers

Definition. A nonempty set S of real numbers is said to be bounded above if there

exists some real number M such that x ≤ M for every x in S. In this case M is said

to be an upper bound for S. Likewise, S is said to be bounded below if there exists

some number m such that m ≤ x for all x in S; m is called a lower bound for S. We

call S bounded if it is bounded both above and below.

Definition. Let S be a nonempty set of real numbers that is bounded above. A supremum or least upper bound of S, denoted sup S, is a real number µ such that 2

1. x ≤ µ, for all x in S;

2. If M is an upper bound for S, then µ ≤ M.

Definition. Let S be a nonempty set of real numbers that is bounded below. An infimum or greatest lower bound of S, denoted inf S, is a real number ν such that

1. ν ≤ x, for all x in S;

2. If m is any lower bound for S, then ν ≥ m.

Axiom (The Completeness Axiom for R). If S is a nonempty set of real numbers

that is bounded above, then sup S exists in R.

1.2 Neighborhoods and Limit Points

Definition. Let x be any real number. The neighborhood of x with positive radius r

is the set

N(x; r) = {y in R : |y − x| < r}

Definition. A deleted neighborhood of x in R, denoted N 0(x; r), is the neighborhood N(x; r) with the point x itself removed.

Definition. Given a nonempty set S in R, a point x in R is said to be a limit point of S if, for each ε > 0, the deleted neighborhood N(x; ε) contains at least one point

of S.

Theorem (Bolzano-Weierstrass Theorem). If S is a bounded, infinite subset of R,

then S has a limit point in R.

1.3 The Limit of a Sequence

Definition. The point c is a cluster point of the sequence {xk} if, for every ε > 0

and every k in N, there is a k1 > k such that xk1 is in N(c; ε). 3

Definition. The sequence {xk} converges to x0 and we say x0 is the limit of {xk} if,

for each neighborhood N(x0; ε), there exists an index k0 such that, whenever k > k0,

xk is in N(x0; ε). We write limk→∞ xk = x0. If a sequence {xk} fails to converge, for whatever reason, then we say that the sequence diverges.

Theorem. The limit of a convergent sequence in R is unique.

Theorem. A convergent sequence is bounded.

Theorem. If a sequence is unbounded, then it must diverge.

Theorem. A bounded, monotonic sequence of real numbers converges.

Definition. Let {xk} be any sequence. Choose any strictly monotonic increasing

sequence k1 < k2 < k3 < ... of natural numbers. For each j in N, let yj = xkj . The

sequence {yj} = {xkj } is called a subsequence of {xk}.

Theorem. The point c is a cluster point of {xk} if and only if there exists a

subsequence xkj that converges to c.

Theorem. Any bounded sequence has a cluster point.

Theorem. A sequence xkj converges to x0 if and only if every subsequence of xkj

converges to x0.

Theorem. A bounded sequence converges if and only if it has exactly one cluster

point. 4

Chapter 2

Stolz-Ces`aroTheorems and the

Stirling Formula

In this chapter, we discuss powerful criteria for convergence of sequences, all due to

Otto Stolz and Ernesto Ces`aro.

2.1 The Stolz-Ces`aroTheorem

Let (a ) and (b ) be real sequences such that (b ) is strictly increasing n n∈N0 n n∈N0 n n∈N0 with limn→∞ bn = ∞. Then, we have

a − a a a a − a lim inf n n−1 ≤ lim inf n ≤ lim sup n ≤ lim sup n n−1 . n→∞ bn − bn−1 n→∞ bn n→∞ bn n→∞ bn − bn−1

In particular a − a a lim n n−1 = lim n , n→∞ bn − bn−1 n→∞ bn

provided that the limit on the left-hand side exists. 5

2.2 The Stolz-Ces`aroTheorem

(Equivalent Formulation)

Let (a ) and (b ) be real sequences such that b > 0, n ∈ and n n∈N0 n n∈N0 n N0

limn→∞ bn = ∞. Then, we have

a a + ··· + a a + ··· + a a lim inf n ≤ lim inf 1 n ≤ lim sup 1 n ≤ lim sup n . n→∞ bn n→∞ b1 + ··· + bn n→∞ b1 + ··· + bn n→∞ bn

In particular a a + ··· + a lim n = lim 1 n , n→∞ bn n→∞ b1 + ··· + bn

provided that the limit on the left-hand side exists.

Proof. This follows directly from the previous by the substitution

an 7→ a0 + ··· + an and bn 7→ b0 + ··· + bn, n ∈ N0.

Letting bn = n, n ∈ N, we obtain the following special cases valid for any real sequence (a ) : n n∈N

a1 + ··· + an a1 + ··· + an lim inf an ≤ lim inf ≤ lim sup ≤ lim sup an, n→∞ n→∞ n n→∞ n n→∞ and

an an lim inf (an − an−1) ≤ lim inf ≤ lim sup ≤ lim sup (an − an−1) . n→∞ n→∞ n n→∞ n n→∞

In particular,

a1 + ··· + an lim an = lim , n→∞ n→∞ n

and

an lim (an − an−1) = lim , n→∞ n→∞ n

provided that the limits on the left-hand sides exist. 6

We call these the additive Stolz-Ces`aro formulas.

Let (a ) be a real sequence with positive members. For n ∈ , let b = log (a ), n n∈N N n 2 n

bn or equivalently, an = 2 . Applying the exponential identities, we obtain

b1+···+bn √ n 2 n = a1 . . . an.

Using the monotonicity of the exponentiation, and the Stolz-Ces`arolimit formulas

above for the sequence (b ) , we obtain n n∈N

√ √ n n lim inf an ≤ lim inf a1 . . . an ≤ lim sup a1 . . . an ≤ lim sup an, n→∞ n→∞ n→∞ n→∞

and

a √ √ a n n n n lim inf ≤ lim inf an ≤ lim sup an ≤ lim inf . n→∞ an−1 n→∞ n→∞ n→∞ an−1

In particular

√ n lim an = lim a1 . . . an, n→∞ n→∞

and

a √ n n lim = lim an, an−1 n→∞

provided that the limits on the left-hand sides exist.

We call these the multiplicative Stolz-Ces`aro formulas. 7

2.3 The Stirling Formula

An alternative approach for deriving some of the limits that follow in Chapter 3 is

to first derive the Stirling Formula. However, the Stirling Formula is much more

involved, compared to the more elementary Stolz-Ces`aroformulas, and would require

elaborate proofs that are not within the scope of this work. 8

Chapter 3

Examples of Sophisticated Limits

3.1 Preliminaries

We begin with a preliminary example:

n (1) lim √ = e. n→∞ n n!

n Indeed, letting an = n!/n , n ∈ N, we calculate

n n an+1 (n + 1)! n n 1 1 lim = lim n+1 · = lim n = lim n = , n→∞ an n→∞ (n + 1) n! n→∞ (n + 1) n→∞ (1 + 1/n) e where we use the Euler’s limit relation. Now, the multiplicative Stolz-Cez`arotheorem

gives

√ n 1/n n! 1 lim an = lim = . n→∞ n→∞ n e

(1) follows.

As a direct application, we have the following example (Lalescu Sequence):

 √  1 lim n+1p(n + 1)! − n n! = . n→∞ e 9

Once again, an easy application of the Stolz-Ces`aroTheorem gives

√ √  √  n+1p(n + 1)! − n n! n n! 1 lim n+1p(n + 1)! − n n! = lim = lim = , n→∞ n→∞ (n + 1) − n n→∞ n e where we used (1).

3.2 Example 1

We are now ready to state our first limit.

Let (a ) be a real sequence with positive terms: 0 < a ∈ , n ∈ . We have the n n∈N n R N following implication

! a − a a a L (2) lim n+1 n = L > 0 ⇒ lim n+1 − √n = e · n→∞ n n→∞ n+1p(n + 1)! n n! 2

2 An important special case is an = n as follows. B˘atinet¸u-Giurgiu Sequence:

! (n + 1)2 n2 lim − √ = e. n→∞ n+1p(n + 1)! n n!

(n+1)2−n2 2n+1 This follows from (2), since limn→∞ n = limn→∞ n = 2.

We now turn to the proof of (2). Let (a ) as above. We first use the Stolz-Ces`aro n n∈N Theorem to the effect that

a a − a a − a n L (3) lim n = lim n+1 n = lim n+1 n · = , n→∞ n2 n→∞ (n + 1)2 − n2 n→∞ n 2n + 1 2 where we used the assumption in (2).

We take the common denominator inside the second limit in (2), and use (1) and (3) 10

to calculate

! √ ! a a a a n n! lim n+1 − √n = lim √n n+1 · − 1 n+1p n n n+1p n→∞ (n + 1)! n! n→∞ n! an (n + 1)!

√ ! n a a n n! = lim √ · n · n n+1 · − 1 n 2 n+1p n→∞ n! n an (n + 1)!

√ ! L a n n! = e · · lim n n+1 · − 1 . n+1p 2 n→∞ an (n + 1)!

It remains to calculate the last limit, which we write as

lim n (bn − 1) = 1, n→∞ where √ n an+1 n! bn = · . n+1p an (n + 1)!

First, note that

√ n an+1 n! lim bn = lim · n+1p n→∞ n→∞ an (n + 1)!

√ a n n! n + 1 n + 1 = lim n+1 · · · n+1p n→∞ an n (n + 1)! n

a n2 (n + 1)2 n + 1 n + 1 1 = lim n+1 · · · · = 1 · · e = 1, 2 2 n+1p n→∞ (n + 1) an n (n + 1)! n e where we used (3) again.

By continuity of the natural logarithm, we obtain

lim ln bn = 0. n→∞ 11

Returning to the main line, we have

eln bn − 1 ln bn
lim n (bn − 1) = lim n e − 1 = lim n ln bn · = lim n ln bn, n→∞ n→∞ n→∞ ln bn n→∞ since eln bn − 1 ex − 1 ex lim = lim = lim = 1. n→∞ ln bn n→∞ x n→∞ 1

Finally, we use the explicit formula above for bn, n ∈ N, and obtain

√ n ! an+1 n! lim n ln bn = lim n ln · n+1p n→∞ n→∞ an (n + 1)!

 a ln n! (n + 1)! = lim n ln n+1 + − . n→∞ an n n + 1

For the first term in the last parentheses, we calculate

a a   lim n ln n+1 = lim n ln n+1 − 1 + 1 n→∞ an n→∞ an

a  = lim n n+1 − 1 n→∞ an

n2 a − a 2 = lim · n+1 n = · L = 2. n→∞ an n L

where we used limn→∞ an+1/an = 1 along with L’Hˆopital’sRule:

   ln an+1 − 1 + 1 an ln (x + 1) lim a = lim = 1. n→∞ n+1 x→0 x an−1

For the last two terms in the parentheses above, we use the Stolz-Cez`aroTheorem

again, and calculate 12

ln n! ln (n + 1)! lim n − n→∞ n n + 1

(n + 1) ln n! − n ln (n + 1)! = lim n→∞ n + 1

(n + 1) (ln 1 + ln 2 + ··· + ln n) − n (ln 1 + ln 2 + ··· + ln n + ln (n + 1)) = lim n→∞ n + 1

ln n! − n ln (n + 1) = lim n→∞ n + 1

(ln (n + 1)! − (n + 1) ln (n + 2)) − (ln n! − n ln (n + 1)) = lim n→∞ (n + 2) − (n + 1)

n + 1 = lim (n + 1) ln n→∞ n + 2

! n + 2n+1 = − ln lim n→∞ n + 1

!  1 n+1 = − ln lim 1 + = − ln e = −1. n→∞ n + 1

Finally, putting everything together, we obtain

lim n (bn − 1) = 2 − 1 = 1 n→∞

Now (2) follows. 13

3.3 Example 2

We have

αn ln ·n! (1) lim n = e, αn = 2 , n→∞ n · Hn

1 1 th 1 where Hn = 1 + 2 + ··· + n , is the n partial sum of the harmonic series.

To derive (1), we first prove the following two lemmas.

Lemma 1.

nn nn (2) < n! < n · , n ≥ 2. e e

Proof. First, we show

 1 n  1 n+1 (3) 1 + < e < . n n

1 Indeed, by the trivial estimate of the integral, since y = x is strictly decreasing, we have 1 Z n+1 dx 1 < < . 1 + n n x n

This gives 1 n + 1 1 < ln (n + 1) − ln n = ln < . n + 1 n n

Hence n + 1n+1  1 n+1 e < = 1 + , n n

and  1 n n + 1n 1 + = < e. n n

Thus, (3) follows.

1 It is well-known that limn→∞ Hn = ∞. 14

We now proceed to prove (2) by induction, with respect to n ≥ 2.

Initial Case:

Let n = 2, then

22 22 e · < 2! < 2 · e e e

4 4 e · < 2 < 2 · e · e2 e2

4 8 < 2 < e e

2 < e < 4.

For the general induction step n ⇒ n + 1, we assume that (2) holds for n.

Therefore, we have

n + 1n+1 1 (n + 1)n+1 nn (n + 1) > (n + 1) · · n · e e nn+1 e

1  1 n+1 1  1 n+1 > (n + 1) · · 1 + · n! = · 1 + · (n + 1) > (n + 1)! e n e n

where 1  1 n+1  1 n+1 · 1 + > 1 and 1 + > e. e n n

Lemma 2.

ln n < Hn < 1 + ln (n + 1) , 1 ≤ n ∈ N.

Proof. Once again, the trivial estimate of the integral gives 15

1 1 Z n+1 dx 1 + + ··· + > = ln (n + 1) 2 n n x

and 1 1 Z n dx + ··· + < = ln n. 2 n 1 x

Combined together, these give

(4) ln (n + 1) < Hn < 1 + ln n, n > 1.

We are now ready to show (1). Taking the natural logarithm of (2), we obtain

n · (ln n − 1) < ln n! < n · (ln n − 1) + ln n.

2 Multiplying through the chain of inequalities by ln n/ (n · Hn), we get

(ln n − 1) · ln n ln n · ln n! [n · (ln n − 1) + ln n] · ln n 2 < 2 < 2 , n ≥ 2. Hn n · Hn n · Hn

2 We replace the lower and upper bounds in (4) with Hn, and obtain

(ln n − 1) · ln n ln · ln n! n · (ln n − 1) + ln n 2 < 2 < , n ≥ 2. [1 + ln (n + 1)] n · Hn n · ln n

For the limit of the lower bound, as n → ∞, we calculate

ln n (ln n − 1) · ln n 1 − n lim 2 = lim 2 = 1, n→∞ [1 + ln (n + 1)] n→∞ h 1+ln(n+1) i ln n

where we used the elementary limits

lim ln n/n = 0 and lim ln (n + 1) / ln(n) = 1, n→∞ n→∞ 16 where both limit relations follows from L’Hˆopital’sRule.

For the upper bound, we have

n · (ln n − 1) + ln n  1 1  lim = lim 1 − + = 1. n→∞ n · ln n n→∞ ln n

Finally, by the Squeeze Theorem, we obtain

ln n · ln n! lim 2 = 1. n→∞ n · Hn

Therefore,

ln n·ln n! α α ln n 2 lim n n = lim e n = lim e n·Hn = e. n→∞ n→∞ n→∞

Example 2 follows. 17

3.4 Example 3

2 Let 0 < x0 < 1 and xn+1 = xn − xn , n ∈ N0. Show that

(1) lim xn = 0; n→∞

(2) lim nxn = 1; n→∞ and find

n (1 − nx ) (3) lim n n→∞ ln n

The first two limit relations are the contents of the next two lemmas.

Lemma 3. The sequence (x ) is decreasing and we have (1). n n∈N0

Proof. First, we use induction to show that xn ∈ (0, 1) for all n ∈ N0. This holds

for n = 0 by assumption. For the general induction step n ⇒ n + 1, if xn ∈ (0, 1),

2 then xn+1 = xn − xn = xn (1 − xn) ∈ (0, 1) clearly holds.

Note that this immediately implies that the sequence (x ) is decreasing, since n n∈N0 2 xn+1 = xn − xn < xn for all n ∈ N0.

Finally, since (x ) is decreasing and bounded from below, the Bolzano-Weierstrass n n∈N0

Theorem applies (see Sec. 1.2), yielding the existence of the limit limn→∞ xn = l. The recurrence relation defining the sequence now gives l = l − l2. Hence l = 0, and

the lemma follows.

Lemma 4. We have the limit relation (2):

lim nxn = 1. n→∞ 18

Proof. We write the limit in (2) as

n (n + 1) − n lim nxn = lim = lim , n→∞ n→∞ 1 n→∞ 1 − 1 xn xn+1 xn

where, in the last equality, we used the additive Stolz-Ces`aroTheorem with an = n and ln = 1 . Note that, by Lemma 3, we have lim 1 = ∞, so that the additive xn n→∞ xn Stolz-Ces`arotheorem holds.2

We calculate (n + 1) − n xn · xn+1 lim 1 1 = lim n→∞ − n→∞ xn − xn+1 xn+1 xn

2 xn · (xn − xn) = lim 2 n→∞ xn

= lim (1 − xn) = 1. n→∞

Lemma 4 follows.

Finally, we turn to the proof of limit relation (3) as follows.

1 − n 1 − n n (1 − nxn) xn xn lim = lim nxn · = lim , n→∞ ln n n→∞ ln n n→∞ ln n where we used Lemma 4. We rewrite the last limit using the additive Stolz-Ces`aro

Theorem, and calculate

2In what follows, we will use the Stolz-Ces`aroTheorem frequently without explicit reference to Sec.2.1. 19

[ 1 − (n + 1)] − [ 1 − n] lim xn+1 xn n→∞ ln (n + 1) − ln n

1 − 1 − 1 = lim xn+1 xn n→∞ 1
ln 1 + n

 1 1  = lim n · − − 1 n→∞ xn · (1 − xn) xn

1 − (1 − x ) − x · (1 − x ) = lim n · n n n n→∞ xn · (1 − xn)

n · x = lim n = 1. n→∞ 1 − xn

We obtain limit relation in (3). 20

3.5 Example 4

In the previous examples, we have seen that the Stolz-Ces`aroTheorems serve as a

powerful tool for calculating difficult limits. However, in some instances, L’Hˆopital’s

Rule provides a short and more efficient approach for deriving certain limits, as we will see in the example below.

Show that

" n #n 1 + 1
1 (1) lim n =√ n→∞ e e

We will derive a more general continuous limit:

1 " 1 1 # x (1 + ) x 1 (2) lim x =√ . x→0+ e e

To do this, we take the natural logarithm of the limit and calculate

 1  1 ! x 1 (1 + x) x 1 (1 + x) x ln  lim  = lim · ln x→0+ e x→0+ x e

1 ln(1 + x) − 1 ln(1 + x) − x = lim x = lim x→0+ x x→0+ x2

1 − 1 1 − (1 + x) 1 = lim 1+x = lim = − . x→0+ 2x x→0+ 2x(1 + x) 2

where, in the last but one equality, we used L’Hˆopital’sRule.

Exponentiating (2), and hence (1) follows. 21

3.6 Example 5

Show that

 r r  xn+1 xn a (1) lim n+1 − n = , n→∞ yn+1 yn b · e

where xn+1 (xn)n≥1, xn > 0; lim 2 = a > 0 n→∞ n xn

yn+1 (yn)n≥1, yn > 0; lim = b > 0. n→∞ nyn

q Let a = n xn . We start by deriving lim an = a . Applying the Stolz-Ces`aro n yn n→∞ n b·e Theorem, we obtain

√ √ √ n n 1 n pn xn xn n xn 2 xn 2n lim √ = lim √ = lim n √ = lim n n→∞ n n y n→∞ n2 n y n→∞ 1 n n→∞ pn yn n n n yn nn

2n xn+1 yn+1 n xn+1 2(n+1) ÷ n+1 2(n+1) · (n+1) (n+1) (n+1) xn = lim = lim n n→∞ xn yn n→∞ n yn+1 2n ÷ n n+1 · n n (n+1) yn

2n 2 n
xn+1 1 n xn+1 · 2 2 · 2 n+1 (n+1) ·xn e2 (n+1) n ·xn 1 a = lim n = · lim y = · . n→∞ n
· yn+1 1 n→∞ n · n+1 e b n+1 (n+1)·yn e n+1 n·yn

This gives

a  a n n + 1 a be lim n+1 = n+1 · · = · · 1 = 1 n→∞ an n + 1 an n be a

and n n a   x ny r y  n+1 a be n+1 n+1 n n n lim = lim 2 · · n = · = e. n→∞ an n→∞ n xn yn+1 xn b a 22

Putting everything together, we obtain:

 rx rx   a  n+1 n+1 n n n+1 lim − = lim (an+1 − an) = lim an · − 1 n→∞ yn+1 yn n→∞ n→∞ an

 a  ln( n+1 −1)  n an e an an+1 a a = lim  ·   · ln  = · 1 · ln e = . n→∞ n ln an+1 an b · e b · e an

Thus, (1) follows. 23

3.7 Example 6

Show that

n! (1) lim (1 + e − Fn) = 1 n→∞

and

(n+1)! (2) lim (1 + e − Fn) = e , n→∞

Pn 1 where Fn = k=0 k! with limn→∞ Fn = e.

We first take the natural logarithm of (1) and calculate

 n! ln lim (1 + e − Fn) = lim n! · ln(1 + e − Fn) n→∞ n→∞

ln(1 + e − Fn) = lim n! · (e − Fn) n→∞ e − Fn

ln(1 + e − Fn) = lim · lim n! · (e − Fn). n→∞ e − Fn n→∞

We claim that the first limit on the right-hand side is equal to 1. Indeed, we let

e − Fn = u → 0, as n → ∞ and apply L’Hˆopital’sRule (to the continuous limit), and obtain

ln(1 + e − F ) ln(1 + u) 1 lim n = lim = lim 1+u = 1. n→∞ e − Fn u→0+ u u→0+ 1

For the remaining (second) limit, we use the Stolz-Ces`aroTheorem, and calculate 24

e − Fn (e − Fn − (e − Fn−1)) lim n! · (e − Fn) = lim = lim n→∞ n→∞ 1 n→∞ 1 1 n! n! − (n−1)!

F − F − 1 1 = lim n−1 n = lim n! = lim n = 0. n→∞ 1 1
n→∞ 1 1
n→∞ 1 (n−1)! · n − 1 (n−1)! · n − 1 1 − n

Hence, (1) follows.

The computations leading to (2) are almost verbatim to that of (1). To derive

(2), the only change is to replace n! by (n + 1)!. For the last limit above, we obtain

1 e − Fn − n! lim (n + 1)!(e − Fn) = lim = lim = 1. n→∞ n→∞ 1 n→∞ 1 1
(n+1)! n! · n+1 − 1

Thus, (2) follows. 25

Bibliography

[1] Steven A. Douglass. Introduction to Mathematical Analysis, Addison-Wesley, Reading, Massachusetts, 1996.

[2] Virgil Nicula. Very Nice and Difficult Limits. The Art of Problem Solving, October 16, 2010.