Question 8.T.16: (Integral Test) Suppose f : [1, ∞) → [0, ∞) is decreasing an...
(Integral Test)
Suppose f : [1, ∞) → [0, ∞) is decreasing and Riemann integrable on [1, b] for all b > 1. Then
(i) The series \Sigma^{∞}_{n=1} f (n) s convergent if and only if the improper integral \int_{1}^{∞}{f (t)} dt is convergent.
(ii) In case the series is convergent, we have
\int_{n+1}^{∞}{f (t)} dt ≤ \sum\limits_{k=n+1}^{∞}{f (k)} ≤ \int_{n}^{∞}{f (t)} dt, n ∈ \mathbb{N}.
(i) Since f is decreasing and integrable over [k − 1, k] ,we have
f (k) ≤ \int_{k-1}^{k}{f (t)} dt ≤ f (k − 1) . (8.24)
Summing up from k = 2 to k = n, we obtain
\sum\limits_{k=2}^{n}{f (k)} ≤ \int_{1}^{n}{f (t)} dt ≤ \sum\limits_{k=1}^{n-1}{f (k)} .
The convergence of the series ensures that the increasing sequence of integrals \int_{1}^{n}{f (t)} dt is bounded above by \Sigma^{∞}_{k=1} f (k) and is therefore convergent. On the other hand, if the improper integral \int_{1}^{∞}{f (t)} dt exists, then it becomes an upper bound for the partial sums of the series, and we conclude that the series is convergent.
(ii) By summing over k from n +1 to m in (8.24) we see that
\sum\limits_{k=n+1}^{m}{f (k)} ≤ \int_{n}^{m}{f (t)} dt ≤ \sum\limits_{k=n}^{m-1}{f (k)}
⇒ \int_{n+1}^{m+1}{f (t)} dt ≤ \sum\limits_{k=n+1}^{m}{f (k)} ≤ \int_{n}^{m}{f (t)} dt.
If the series converges then so does the improper integral, and the desired result is obtained by letting m → ∞.