5.6. EXERCISES 135

This follows easily from consideration of the following graph. The larger trapezoidis obtained from the tangent line through (n+ 1

2 , ln(n+12 )).

n n+1n+ 12

(a) Now ∑n−1k=1

(∫ k+1k ln(t)dt− 1

2 (ln(k+1)+ ln(k)))

is an increasing sequence ofpartial sums.

(b) Next consider the following computations coming from the above inequalities

∫ n

1ln(t)dt−

n−1

∑k=1

12(ln(k+1)+ ln(k))

=n−1

∑k=1

(∫ k+1

kln(t)dt− 1

2(ln(k+1)+ ln(k))

)

≤n−1

∑k=1

ln(k+1/2)− 12(ln(k+1)+ ln(k))

=n−1

∑k=1

12(ln(k+1/2)− ln(k))−

n

∑k=2

12(ln(k)− ln(k−1/2))

=12

ln(

32

)− 1

2(ln(n)− ln(n−1/2)) ≤ 1

2ln(

32

)Therefore, the series in part a.) converges to some c.

(c) Note that the series in a.) equals

Sn ≡∫ n

1ln(t)dt−

n−1

∑k=1

12(ln(k+1)+ ln(k)) .

Hence limn→∞ exp(Sn) = ec. Thus

limn→∞

exp(n lnn−n+1)

∏n−1k=1 exp

(√(k+1)k

)=

nne−ne

∏n−1k=1

√k+1∏

n−1k=1

√k

=nne−nen1/2

∏n−1k=1

√k+1∏

nk=1

√k=

nn+(1/2)en!en = ec

Thus there is a constant k = 1/e1−c such that limn→∞n!en

n(n+(1/2)) = k. It is possible

to show that k =√

2π but in most applications, it suffices to know the existenceof the limit. This is Stirling’s formula.