由收斂半徑R = 1,對(duì)任意0 ≤ x < 1,f(x)有定義.
又由a[n] > 0,對(duì)任意0 ≤ x < 1,有a[n]·x^n ≤ a[n],故f(x) = ∑a[n]·x^n ≤ ∑a[n].
另x → 1-,得s = lim{x → 1-} f(x) ≤ ∑a[n].(注:這里不排除∑a[n] = +∞的情形).
對(duì)任意ε > 0與N > 0,存在δ = 1-1/(1+ε)^(1/N) > 0.
當(dāng)1 > x > 1-δ,有x^N > (1-δ)^N = 1/(1+ε).
此時(shí)有∑{0 ≤ n ≤ N} a[n] = 1/x^N·∑{0 ≤ n ≤ N} a[n]·x^N
≤ 1/x^N·∑{0 ≤ n ≤ N} a[n]·x^n
≤ 1/x^N·∑{0 ≤ n} a[n]·x^n
≤ (1+ε)f(x)
≤ (1+ε)s.
即有∑{0 ≤ n ≤ N} a[n] ≤ (1+ε)s對(duì)任意ε > 0與N > 0均成立.
由ε的任意性及s ≥ 0,有∑{0 ≤ n ≤ N} a[n] ≤ s對(duì)任意N > 0均成立.
于是正項(xiàng)級(jí)數(shù)∑a[n]收斂,并成立∑a[n] ≤ s.
綜合得∑a[n] = s.