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Series of Nonnegative Real Numbers, Comparison, and the Geometric Series

A series with nonnegative terms converges exactly when its partial sums are bounded above, and its sum is then their supremum; the comparison test; and the geometric series with its tails.

Statement

In the setting of The Real Numbers: Standing Notation and Background, let (ak)k∈N(a_{k})_{k\in\mathbb{N}} and (bk)k∈N(b_{k})_{k\in\mathbb{N}} be sequences of real numbers, with partial sums sns_{n} and tnt_{n} respectively, and let convergence of a series and its sum be as defined there. Then the following hold.

1. (Criterion for nonnegative terms) Suppose 0≤ak0\le a_{k} for every k∈Nk\in\mathbb{N}. Then sn≤sn+1s_{n}\le s_{n+1} for every n∈Nn\in\mathbb{N}, and ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges if and only if the set {sn:n∈N}\{s_{n}:n\in\mathbb{N}\} is bounded above; in that case

∑k=1∞ak=sup⁡{sn:n∈N}.\sum_{k=1}^{\infty}a_{k}=\sup\{s_{n}:n\in\mathbb{N}\}.

2. (Domination by the sum) Suppose 0≤ak0\le a_{k} for every k∈Nk\in\mathbb{N} and ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges. Then

0≤sn≤∑k=1∞akfor every n∈N.0\le s_{n}\le\sum_{k=1}^{\infty}a_{k}\qquad\text{for every }n\in\mathbb{N}.

3. (Comparison test) Suppose 0≤ak≤bk0\le a_{k}\le b_{k} for every k∈Nk\in\mathbb{N} and that ∑k=1∞bk\sum_{k=1}^{\infty}b_{k} converges. Then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges and

∑k=1∞ak≤∑k=1∞bk.\sum_{k=1}^{\infty}a_{k}\le\sum_{k=1}^{\infty}b_{k}.

4. (Geometric series) Let r∈Rr\in\mathbb{R} satisfy 0≤r<10\le r<1, and let rkr^{k} denote the natural power. Then 1−r1-r is positive, the sequence (rk)k∈N(r^{k})_{k\in\mathbb{N}} converges to 00, and

∑k=1nrk=r−rn+11−rfor every n∈N.\sum_{k=1}^{n}r^{k}=\frac{r-r^{n+1}}{1-r}\qquad\text{for every }n\in\mathbb{N}.

The series ∑k=1∞rk\sum_{k=1}^{\infty}r^{k} converges, and for every n∈Nn\in\mathbb{N}

∑k=1∞rk=r1−r,∑k=1∞rk−∑k=1nrk=rn+11−r.\sum_{k=1}^{\infty}r^{k}=\frac{r}{1-r}, \qquad \sum_{k=1}^{\infty}r^{k}-\sum_{k=1}^{n}r^{k}=\frac{r^{n+1}}{1-r}.

In particular, taking r=12r=\tfrac{1}{2}, the series ∑k=1∞(12)k\sum_{k=1}^{\infty}\bigl(\tfrac{1}{2}\bigr)^{k} converges with sum 11, and ∑k=1∞(12)k−∑k=1n(12)k=(12)n\sum_{k=1}^{\infty}\bigl(\tfrac{1}{2}\bigr)^{k}-\sum_{k=1}^{n}\bigl(\tfrac{1}{2}\bigr)^{k}=\bigl(\tfrac{1}{2}\bigr)^{n} for every n∈Nn\in\mathbb{N}.

5. (Tail bound for a dominated series of nonnegative terms) Let (μk)k∈N(\mu_{k})_{k\in\mathbb{N}} be a sequence of nonnegative real numbers such that ∑k=1∞μk\sum_{k=1}^{\infty}\mu_{k} converges, let M∈RM\in\mathbb{R} be nonnegative, and let (wk)k∈N(w_{k})_{k\in\mathbb{N}} be a sequence of real numbers with 0≤wk≤M0\le w_{k}\le M for every k∈Nk\in\mathbb{N}. Then the series ∑k=1∞μkwk\sum_{k=1}^{\infty}\mu_{k}w_{k} converges and, for every n∈Nn\in\mathbb{N},

0≤∑k=1∞μkwk−∑k=1nμkwk≤M(∑k=1∞μk−∑k=1nμk).0\le\sum_{k=1}^{\infty}\mu_{k}w_{k}-\sum_{k=1}^{n}\mu_{k}w_{k}\le M\Bigl(\sum_{k=1}^{\infty}\mu_{k}-\sum_{k=1}^{n}\mu_{k}\Bigr).

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