168 CHAPTER 6. INFINITE SERIES

15. Show that if {pi}∞

i=1 are the prime numbers, then ∑∞i=1

1pi= ∞. That is, there are

enough primes that the sum of their reciprocals diverges. Hint: Let π (n) denote thenumber of primes less than equal to n,

{p1, ..., pπ(n)

}. Then explain why

n

∑k=1

1k≤

(n

∑k=1

1pk

1

)· · ·

(n

∑k=1

1pk

π(n)

)≤

π(n)

∏k=1

11− 1

pk

≤π(n)

∏k=1

e2/pk = e2∑π(n)k=1

1pk

and consequently why limn→∞ π (n) = ∞ and ∑∞i=1

1pi= ∞.

6.8 Series of FunctionsInfinite sequences of functions were discussed earlier. Remember, there were two kindsof convergence, pointwise and uniform. As was just done for series of numbers, once youunderstand sequences, it is no problem to consider series. In this case, series of functions.

Definition 6.8.1 Let { fn} be a sequence of functions defined on D. Then(∞

∑k=1

fk

)(x)≡ lim

n→∞

n

∑k=1

fk (x) (6.8)

whenever the limit exists. Thus there is a new function denoted by

∞

∑k=1

fk (6.9)

and its value at x is given by the limit of the sequence of partial sums in 6.8. If for all x ∈ D,the limit in 6.8 exists, then 6.9 is said to converge pointwise. ∑

∞k=1 fk is said to converge

uniformly on D if the sequence of partial sums,{∑nk=1 fk}∞

n=1 converges uniformly. If theindices for the functions start at some other value than 1, you make the obvious modificationto the above definition as was done earlier with series of numbers.

Theorem 6.8.2 Let { fn} be a sequence of functions defined on D. The series ∑∞k=1 fk

converges pointwise if and only if for each ε > 0 and x ∈ D, there exists Nε,x which may

depend on x as well as ε such that when q > p ≥ Nε,x,∣∣∣∑q

k=p fk (x)∣∣∣< ε . The series ∑

∞k=1 fk

converges uniformly on D if for every ε > 0 there exists Nε such that if q > p ≥ Nε then

supx∈D

∣∣∣∣∣ q

∑k=p

fk (x)

∣∣∣∣∣< ε (6.10)

Proof: The first part follows from Theorem 6.1.8. The second part follows from ob-serving the condition is equivalent to the sequence of partial sums forming a uniformlyCauchy sequence and then by Corollary 4.9.5, these partial sums converge uniformly to afunction which is the definition of ∑

∞k=1 fk.

Is there an easy way to recognize when 6.10 happens? Yes, there is. It is called theWeierstrass M test.

Theorem 6.8.3 Let { fn} be a sequence of functions defined on D. Suppose thereexists Mn such that sup{| fn (x)| : x ∈ D}< Mn and ∑

∞n=1 Mn converges. Then ∑

∞n=1 fn con-

verges uniformly on D.