Let f be positive and nonincreasing on [k, ∞) , where k is a positive integer,

8.4.5. Let f be positive and nonincreasing on [k, ∞) , where k is a positive integer,

and let F be any antiderivative of f . Prove that

(i) the series ∞ ∑ n =k f (n) converges if and only if F is bounded above on [k, ∞) ; (ii) if lim x → ∞ F (x) = 0 , then ∞ ∑ n =k f (n) converges. Furthermore, if S is the sum of ∑ ∞

n =k f (n) , then

−F(m + 1) ≤ ∑ f (n) ≤ −F(m) for m ≥k

n =m+1

f (k) − F(k + 1) ≤ S ≤ f (k) − F(k), (8.2) and

0 ≤S− ∑ f (n) − F(m + 1) ≤m for m ≥ k.

n =k

Solution. (i) For n ≥ k we apply the mean value theorem to F on [n,n + 1]. Therefore

F (n + 1) − F(n) = F ′ (x n ) = f (x n ),

for some x n ∈ (n,n + 1). Since f is nonincreasing, we have

(8.4) But

f (n + 1) ≤ f (x n ) = F(n + 1) − F(n) ≤ f (n) for n ≥ k.

∑ (F(n + 1) − F(n)) = F(m + 1) − F(k) for m ≥ k,

n =k

so relation (8.4) implies

∑ f (n + 1) ≤ F(m + 1) − F(k) ≤ ∑ f (n) for n ≥ k.

Now, if ∞ ∑ n =k f (n) converges, the right-hand term in (8.5) is bounded above, so that (8.5) implies that F is bounded on the set of integers that are greater than or equal to k. But F ′ (x) = f (x) > 0, and consequently F is increasing on [k, ∞); hence

F is bounded on [k, ∞). Conversely, if F is bounded on [k, ∞), relation (8.5) implies that the sequence of partial sums for the series ∞

∑ ∞ n =k f (n + 1) is bounded above. Thus, ∑ n =k f ∞ (n + 1) and ∑ n =k f (n) are convergent.

(ii) The requirement that lim x → ∞ F (x) = 0 can always be met if f has an antideriva- tive that is bounded above. For example, if f (x) = 1/(1 + x 2 ) we would get

F (x) = arctan x − π /2 instead of arctanx.

8.4 Qualitative Results 323 We introduce the following notation:

S m = ∑ f (n) , E m = ∑ f (n) ;

n =k

n =m+1

hence E m is the error if we approximate S by S m . If lim ∞ F (x) = 0, then F is bounded on [k, ∞), so that the series ∑ x ∞ → n =k f (n) is convergent, by part (i). Now, if we let m →∞ in (8.5), we obtain

∑ f (n + 1) ≤ −F(k) ≤ ∑ f (n) .

n =k

n =k

Since (8.5) is valid when any k ′ ≥ k is substituted for k, we have

∑ f (n + 1) ≤ −F(m) ≤ ∑ f (n) for all m ≥ k.

The left side of (8.6) implies that E m ≤ −F(m), and the right side of (8.6) implies −F(m + 1) ≤ E m . Thus, inequality (8.1) of (ii) holds for m ≥ k. Inequality (8.2) follows from (8.1); just let m = k in (8.1) and add f (k) throughout, noticing that S =E k + f (k). Moreover, relation (8.1) implies

0 ≤E m + F(m + 1) = (S − S m ) + F(m + 1) ≤ F(m + 1) − F(m) for m ≥ k; hence (8.4) yields

0 ≤ S − (S m − F(m + 1)) ≤ F(m) for m ≥ k,

which is inequality (8.3).

E XAMPLE

1. The series ∞ ∑ n =1 n −p converges if p > 1 and diverges if p < 1. Indeed, since F (x) = x 1 −p /(1 − p) for x ≥ 1, then F(x)→0 as x→ + ∞ if p > 1, and

F (x)→ + ∞ as x→ + ∞ if p < 1. Thus, for any p > 1,

(8.3) implies 0 ≤ S − (S −p − F(m + 1)) ≤ m for m ≥ 1.

4 2 approximates ∑ n =1 n e −n accu- rately to four places. Indeed, since F √ (x) = −e −x /2 , then −F(m) < 0.5/10 4 when m ≥4>2 ln 10. ⊓ ⊔

E 2 XAMPLE 2. Inequality (8.1) implies that S ∞

324 8 Antiderivatives