3.4 Convergence results for series with non-negative terms

We focus here on series with non-negative terms. This simplifies the situation significantly, since the sequence of partial sums is non-decreasing. This gives us access to useful tools such as the monotone converge theorem.

The boundedness and comparison tests

We begin by converting the monotone convergence theorem into the language of series.

Lemma 3.26 (Boundedness test for series).

Suppose (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} is a non-negative sequence. Then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges if and only if the sequence of partial sums (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is bounded.

In other words, given a sequence (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} with ak≥0a_{k}\geq 0 for all k∈ℕk\in\mathbb{N}, the boundedness test for series says that the following are equivalent:

  1. 1.
    ​

    ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges;

  2. 2.
    ​

    There exists M≥0M\geq 0 such that 0≤∑k=1nak≤M0\leq\sum_{k=1}^{n}a_{k}\leq M for all n∈ℕn\in\mathbb{N}.

Proof.

If ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges, then by definition the sequence of partial sums (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is convergent and hence bounded by the boundedness test for sequences from Lemma 2.49.

Conversely, assume that (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is bounded. Since the terms aka_{k} are non-negative, the sequence (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is monotone; in particular, sn≤sn+1s_{n}\leq s_{n+1} for all n∈ℕn\in\mathbb{N}. By the monotone convergence theorem, the sequence (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} must therefore be convergent as it is monotone and bounded. ∎

The boundedness test (which, at heart, is just a restatement of the monotone convergence theorem for sequences) will form the backbone of all our convergence results for series with non-negative terms. However, it will take some ingenuity to find different ways in which to apply it. We first observe the following simple corollary, which helps us relate convergence properties of different series.

Corollary 3.27 (Comparison test).

Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} and (bk)k∈ℕ(b_{k})_{k\in\mathbb{N}} satisfy 0≤ak≤bk0\leq a_{k}\leq b_{k} for all k∈ℕk\in\mathbb{N}.

  1. 1
    ​

    If ∑k=1∞bk\sum_{k=1}^{\infty}b_{k} converges, then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges.

  2. 2
    ​

    If ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} diverges, then ∑k=1∞bk\sum_{k=1}^{\infty}b_{k} diverges.

Proof.

We shall only prove part 1 since part 2 is an equivalent form of part 1 (it is the contrapositive).

Suppose (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} and (bk)k∈ℕ(b_{k})_{k\in\mathbb{N}} satisfy the hypotheses of part 1 of the corollary. Since ∑k=1∞bk\sum_{k=1}^{\infty}b_{k} converges, by the boundedness test from Lemma 3.26 the sequence of partial sums of this series is bounded. In particular, there exists some M>0M>0 such that

0≤∑k=1nbk≤Mfor all n∈ℕ,0\leq\sum_{k=1}^{n}b_{k}\leq M\qquad\text{for all $n\in\mathbb{N}$,}

where we have also used the hypothesis that bk≥0b_{k}\geq 0 for all k∈ℕk\in\mathbb{N}. Since 0≤ak≤bk0\leq a_{k}\leq b_{k} for all k∈ℕk\in\mathbb{N}, we therefore see that

0≤∑k=1nak≤∑k=1nbk≤Mfor all n∈ℕ,0\leq\sum_{k=1}^{n}a_{k}\leq\sum_{k=1}^{n}b_{k}\leq M\qquad\text{for all $n\in% \mathbb{N}$,}

which means that the sequence of partial sums of the series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} is also bounded. We may therefore apply the boundedness test from Lemma 3.26 once again (this time using the reverse implication) to conclude that ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges. ∎

Exercise 3.28.

Give a direct proof of part 2 of Corollary 3.27, using an argument similar to that used for part 1.

Example 3.29.

The series ∑k=1∞1k2\sum_{k=1}^{\infty}\frac{1}{k^{2}} converges.

Proof.

Since k⁢(k+1)=k2+k≤2⁢k2k(k+1)=k^{2}+k\leq 2k^{2}, we have

0≤1k2=22⁢k2≤2k⁢(k+1)for all k∈ℕ.0\leq\frac{1}{k^{2}}=\frac{2}{2k^{2}}\leq\frac{2}{k(k+1)}\qquad\text{for all $% k\in\mathbb{N}$.}

Recall from Example 3.14 that ∑k=1∞1k⁢(k+1)\sum_{k=1}^{\infty}\frac{1}{k(k+1)} is a convergent telescoping series and so, by the limit laws, ∑k=1∞2k⁢(k+1)\sum_{k=1}^{\infty}\frac{2}{k(k+1)} converges. Thus, ∑k=1∞1k2\sum_{k=1}^{\infty}\frac{1}{k^{2}} converges by the comparison test. ∎

Exercise 3.30.

Using the comparison test, determine whether the following series converge or diverge.

  1. (i)
    ​

    ∑k=1∞3⁢k+14⁢k3−3\displaystyle\sum_{k=1}^{\infty}\frac{3k+1}{4k^{3}-3};

  2. (ii)
    ​

    ∑k=1∞1(log2⁡(k+1))k\displaystyle\sum_{k=1}^{\infty}\frac{1}{(\log_{2}(k+1))^{k}};

  3. (iii)
    ​

    ∑k=1∞k2k!\displaystyle\sum_{k=1}^{\infty}\frac{k^{2}}{k!}.

Hint: for (ii), note that it is enough to consider the tail ∑k=3∞1(log2⁡(k+1))k\sum_{k=3}^{\infty}\frac{1}{(\log_{2}(k+1))^{k}}. What can you say about log2⁡(k+1)\log_{2}(k+1) for k≥3k\geq 3?

We can use the comparison test to prove a converse to Theorem 3.10, showing that any sequence of decimal digits describes a real number.

Exercise 3.31.

Let (dk)k∈ℕ(d_{k})_{k\in\mathbb{N}} be a sequence with dk∈{0,1,…,9}d_{k}\in\{0,1,\dots,9\} for all k∈ℕk\in\mathbb{N}. Show that ∑k=1∞dk⋅10−k\sum_{k=1}^{\infty}d_{k}\cdot 10^{-k} converges to some number in x∈[0,1]x\in[0,1]. Conclude that x=0.d1⁢d2⁢d3⁢…x=0.d_{1}d_{2}d_{3}\ldots.

The ratio test

There are a number of inventive ways the comparison test can be applied to derive new convergence tests. Here we consider the ratio test, which follows by combining the comparison test with the facts we know about geometric series.

Theorem 3.32 (Ratio test).

Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} be a positive sequence and suppose the sequence of consecutive ratios (ak+1/ak)k∈ℕ\big{(}a_{k+1}/a_{k}\big{)}_{k\in\mathbb{N}} converges with r:=limk→∞ak+1akr:=\lim_{k\to\infty}\frac{a_{k+1}}{a_{k}}.

  1. 1
    ​

    If 0≤r<10\leq r<1, then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges;

  2. 2
    ​

    If r>1r>1, then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} diverges.

Example 3.33.

The series ∑k=1∞k⁢2−k\sum_{k=1}^{\infty}k2^{-k} converges.

Proof.

If we set ak:=k⁢2−ka_{k}:=k2^{-k} for all k∈ℕk\in\mathbb{N} then (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} is a positive sequence and

ak+1ak=k+12⁢k→12as k→∞.\frac{a_{k+1}}{a_{k}}=\frac{k+1}{2k}\to\frac{1}{2}\qquad\text{as $k\to\infty$.}

Since the limit 1/2<11/2<1, the ratio test implies that ∑k=1∞k⁢2−k\sum_{k=1}^{\infty}k2^{-k} converges. ∎

Remark 3.34.

The ratio test says nothing about the case r=1r=1. For both ∑k=1∞1k2\sum_{k=1}^{\infty}\frac{1}{k^{2}} from Example 3.29 and ∑k=1∞1\sum_{k=1}^{\infty}1 from Example 3.6, the limit of the ratios is 11; however, the former series converges whilst the latter series diverges. Thus, the ratio test is inconclusive in the r=1r=1 case.

Proof (of Theorem 3.32).

We shall only prove part 1; the proof of part 2 is left as an exercise (see Exercise 3.35 below).

Let rk:=ak+1/akr_{k}:=a_{k+1}/a_{k} for all k∈ℕk\in\mathbb{N} so that, by hypothesis, r:=limk→∞rk<1r:=\lim_{k\to\infty}r_{k}<1. Let ε:=1−r2>0\varepsilon:=\frac{1-r}{2}>0. By the ε\varepsilon-NN definition of a limit, there exists N∈ℕN\in\mathbb{N} such that |rn−r|<ε|r_{n}-r|<\varepsilon for all n≥Nn\geq N. Thus, by the triangle inequality,

(3.5) (3.5) rn=|r+rn−r|≤r+|rn−r|<r+ε=1+r2for all n≥N,r_{n}=|r+r_{n}-r|\leq r+|r_{n}-r|<r+\varepsilon=\frac{1+r}{2}\qquad\text{for % all $n\geq N$,}

For any k∈ℕk\in\mathbb{N} with k>Nk>N, we may write

ak=aN⋅aN+1aN⋅⋯⋅ak−1ak−2⋅akak−1=aN⋅rN⋅⋯⋅rk−1.a_{k}=a_{N}\cdot\frac{a_{N+1}}{a_{N}}\cdot\dots\cdot\frac{a_{k-1}}{a_{k-2}}% \cdot\frac{a_{k}}{a_{k-1}}=a_{N}\cdot r_{N}\cdot\dots\cdot r_{k-1}.

Applying (3.5) to each of the factors rnr_{n} for N≤n≤k−1N\leq n\leq k-1, we deduce that

(3.6) (3.6) 0<ak<aN⁢sk−Nwhere s:=1+r2 satisfies 0<s<1. 0<a_{k}<a_{N}s^{k-N}\qquad\text{where $s:=\displaystyle\frac{1+r}{2}$ % satisfies $0<s<1$. }

Since 0<s<10<s<1, we know from Example 3.4 that ∑k=N+1∞aN⁢sk−N\sum_{k=N+1}^{\infty}a_{N}s^{k-N} is a convergent geometric series. In light of (3.6), the series ∑k=N+1∞ak\sum_{k=N+1}^{\infty}a_{k} converges by the comparison test. Finally, since we have shown a tail converges, we conclude that the whole series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges. ∎

Exercise 3.35.

In this exercise we complete the proof of and slightly extend Theorem 3.32.

  1. (i)
    ​

    Prove part 2 of the ratio test, using an argument similar to that used for part 1.

  2. (ii)
    ​

    Adapt the argument used in part (i) to prove the following. Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} be a positive sequence and suppose ak+1ak→∞\frac{a_{k+1}}{a_{k}}\to\infty as k→∞k\to\infty (see Definition 2.53). Then ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} diverges.

Exercise 3.36.

Use the ratio test to determine whether the following series converge or diverge.

  1. (i)
    ​

    ∑k=1∞k3⁢22⁢k3k+1\displaystyle\sum_{k=1}^{\infty}\frac{k^{3}2^{2k}}{3^{k}+1};

  2. (ii)
    ​

    ∑k=1∞5kk!\displaystyle\sum_{k=1}^{\infty}\frac{5^{k}}{k!};

  3. (iii)
    ​

    ∑k=1∞k!5k\displaystyle\sum_{k=1}^{\infty}\frac{k!}{5^{k}};

  4. (iv)
    ​

    ∑k=1∞k10(2⁢k−1)!\displaystyle\sum_{k=1}^{\infty}\frac{k^{10}}{(2k-1)!}.

Divergence of the harmonic series

We now consider the interesting example of the harmonic series ∑k=1∞1/k\sum_{k=1}^{\infty}1/k. This example is so important that we give it its own theorem!

Theorem 3.37 (Harmonic series).

The harmonic series ∑k=1∞1/k\sum_{k=1}^{\infty}1/k diverges.

As with Example 3.16 above, Theorem 3.37 provides an example of a series whose terms converge to zero, but which nevertheless diverges. However, the divergence of the harmonic series is a much more subtle affair. There are many different proofs of this result; here we opt for a neat trick based on grouping together terms of similar magnitude. We shall go on to further explore this idea below, and use it to develop a new convergence test.

Proof (of Theorem 3.37).

Let (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} denote the sequence of partial sums of ∑k=1∞1k\sum_{k=1}^{\infty}\frac{1}{k}, so that

sn=1+12+⋯+1n,for n∈ℕ.s_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n},\qquad\text{for $n\in\mathbb{N}$.}

Our goal is to show that (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is divergent. By the boundedness test from Lemma 3.26, it is enough to show that (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is unbounded.

Consider the subsequence (s2n)n∈ℕ0(s_{2^{n}})_{n\in\mathbb{N}_{0}}. For each n∈ℕ0n\in\mathbb{N}_{0}, we isolate the term 11 in the sum s2ns_{2^{n}} and partition the remaining terms into ‘blocks’ that each end on a term whose denominator is a power of 2:

s2n=1+(12)+(13+14)+(15+16+17+18)+⋯+(12n−1+1+12n−1+2+⋯+12n).s_{2^{n}}=1+\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+% \left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots+\left(% \frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\cdots+\frac{1}{2^{n}}\right).

Note that there are precisely nn blocks (ending on denominators 21,22,…,2n2^{1},2^{2},\ldots,2^{n}). Since the sequence (1/n)n∈ℕ(1/n)_{n\in\mathbb{N}} is decreasing, each term in a given block is bounded below by the last term in that block. From this, we see that

s2n=1+(12)⏟≥1⋅12+(13+14)⏟≥2⋅14+(15+16+17+18)⏟≥4⋅18+⋯+(12n−1+1+12n−1+2+⋯+12n)⏟≥2n−1⋅12n,s_{2^{n}}=1+\underbrace{\left(\frac{1}{2}\right)}_{\displaystyle\geq 1\cdot% \frac{1}{2}}+\underbrace{\left(\frac{1}{3}+\frac{1}{4}\right)}_{\displaystyle% \geq 2\cdot\frac{1}{4}}+\underbrace{\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+% \frac{1}{8}\right)}_{\displaystyle\geq 4\cdot\frac{1}{8}}+\cdots+\underbrace{% \left(\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\cdots+\frac{1}{2^{n}}\right)}_{% \displaystyle\geq 2^{n-1}\cdot\frac{1}{2^{n}}},

where we bound each block below by the number of terms multiplied by the value of the smallest term within the block. In particular,

s2n≥1+12+⋯+12⏟n⁢ blocks=1+n2,s_{2^{n}}\geq 1+\underbrace{\frac{1}{2}+\cdots+\frac{1}{2}}_{n\text{ blocks}}=% 1+\frac{n}{2},

where we have used the fact that the lower bound for each block always equals 1/21/2 and that there are precisely nn blocks in the sum. Thus, the subsequence (s2n)n∈ℕ(s_{2^{n}})_{n\in\mathbb{N}} is unbounded, and therefore the parent sequence (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is also unbounded. ∎

Partial sums of the harmonic series (values are approximate, to two decimal places). Plotting the sequence
nn 1 2 3 4 5 6 7 8 9 10 11 12
∼∑k=1n1k\sim\sum_{k=1}^{n}\frac{1}{k} 1 1.5 1.83 2.08 2.28 2.45 2.59 2.72 2.83 2.93 3.02 3.10
Figure 3.2: Partial sums of the harmonic series (values are approximate, to two decimal places). Plotting the sequence (∑k=1n1k)n∈ℕ\big{(}\sum_{k=1}^{n}\frac{1}{k}\big{)}_{n\in\mathbb{N}}, one could be tempted to believe it converges, but it is in fact unbounded!

The partial sums of of the harmonic series are unbounded, but they grow very slowly. As shown in the table in Figure 3.2, we only surpass 33 after summing 1111 terms. If you were to extend the table further, you would find that we only surpass 100100 after summing

15 092 688 622 113 788 323 693 563 264 538 101 449 859 497terms;15\,092\,688\,622\,113\,788\,323\,693\,563\,264\,538\,101\,449\,859\,497\qquad% \text{terms;}

this is more than 1515 million, trillion, trillion, trillion terms!

The condensation test and pp-series

The ideas used to prove the divergence of the harmonic series in Theorem 3.37 can be developed into a general test.

Theorem 3.38 (Condensation test).

Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} be a non-increasing, non-negative sequence. Then the following are equivalent:

  1. 1
    ​

    The series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges;

  2. 2
    ​

    The series ∑k=0∞2k⁢a2k\sum_{k=0}^{\infty}2^{k}a_{2^{k}} converges.

We refer to ∑k=0∞2k⁢a2k\sum_{k=0}^{\infty}2^{k}a_{2^{k}} as the condensed series.

We can revisit the harmonic series ∑k=1∞1/k\sum_{k=1}^{\infty}1/k from Theorem 3.37 in light of Theorem 3.38. Here the sequence of terms (1/k)k∈ℕ(1/k)_{k\in\mathbb{N}} is non-increasing and non-negative, so we may apply the condensation test. The condensed series is ∑k=1∞1\sum_{k=1}^{\infty}1 which clearly diverges (see Example 3.6). Thus, by the condensation test, we again see ∑k=1∞1/k\sum_{k=1}^{\infty}1/k diverges. A further example is given by the following.

Example 3.39.

The series ∑k=2∞1k⁢log2⁡k\sum_{k=2}^{\infty}\frac{1}{k\log_{2}k} diverges, where log2\log_{2} denotes the base 22 logarithm.

Proof.

Since the sequence of terms (1k⁢log2⁡k)k=2∞\big{(}\frac{1}{k\log_{2}k}\big{)}_{k=2}^{\infty} is non-increasing and non-negative, we can apply the condensation test. The condensed series is the harmonic series ∑k=1∞1/k\sum_{k=1}^{\infty}1/k, which we know diverges. Thus, by the condensation test, ∑k=2∞1k⁢log2⁡k\sum_{k=2}^{\infty}\frac{1}{k\log_{2}k} diverges. ∎

We can interpret Example 3.39 as a two-fold application of the condensation test: we first condense ∑k=2∞1k⁢log2⁡k\sum_{k=2}^{\infty}\frac{1}{k\log_{2}k} to the harmonic series ∑k=1∞1/k\sum_{k=1}^{\infty}1/k and then condense the harmonic series to ∑k=1∞1\sum_{k=1}^{\infty}1.

Proof (of Theorem 3.38).

We show 1 ⇒\Rightarrow 2; the reverse implication is left as an exercise (see Exercise 3.40 below). Let

sn\displaystyle s_{n} :=∑k=1nak=a1+a2+a3+a4+⋯+an,\displaystyle:=\sum_{k=1}^{n}a_{k}=a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{n},
tn\displaystyle t_{n} :=∑k=0n−12k⁢a2k=a1+2⁢a2+4⁢a4+8⁢a8+⋯+2n−1⁢a2n−1,for all n∈ℕ,\displaystyle:=\sum_{k=0}^{n-1}2^{k}a_{2^{k}}=a_{1}+2a_{2}+4a_{4}+8a_{8}+% \cdots+2^{n-1}a_{2^{n-1}},\qquad\text{for all $n\in\mathbb{N}$,}

so that (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} and (tn)n∈ℕ(t_{n})_{n\in\mathbb{N}} are sequences of partial sums of the series ∑n=1∞an\sum_{n=1}^{\infty}a_{n} and ∑n=0∞2n⁢a2n\sum_{n=0}^{\infty}2^{n}a_{2^{n}}, respectively. As with the proof of the divergence of the harmonic series in Theorem 3.37, we consider the subsequence (s2n)n∈ℕ(s_{2^{n}})_{n\in\mathbb{N}} and partition the terms of s2ns_{2^{n}} into ‘blocks’

s2n=a1+(a2)+(a3+a4)+(a5+a6+a7+a8)+⋯+(a2n−1+1+a2n−1+2+⋯+a2n).s_{2^{n}}=a_{1}+(a_{2})+(a_{3}+a_{4})+(a_{5}+a_{6}+a_{7}+a_{8})+\cdots+(a_{2^{% n-1}+1}+a_{2^{n-1}+2}+\cdots+a_{2^{n}}).

Since (an)n∈ℕ(a_{n})_{n\in\mathbb{N}} is non-increasing, we have

s2n=a1+(a2)⏟≥1⋅a2+(a3+a4)⏟≥2⋅a4+(a5+a6+a7+a8)⏟≥4⋅a8+⋯+(a2n−1+1+a2n−1+2+⋯+a2n)⏟≥2n−1⋅a2ns_{2^{n}}=a_{1}+\underbrace{(a_{2})}_{\displaystyle\geq 1\cdot a_{2}}+% \underbrace{(a_{3}+a_{4})}_{\displaystyle\geq 2\cdot a_{4}}\\ +\underbrace{(a_{5}+a_{6}+a_{7}+a_{8})}_{\displaystyle\geq 4\cdot a_{8}}+% \cdots+\underbrace{(a_{2^{n-1}+1}+a_{2^{n-1}+2}+\cdots+a_{2^{n}})}_{% \displaystyle\geq 2^{n-1}\cdot a_{2^{n}}}

and therefore

2⁢s2n\displaystyle 2s_{2^{n}} ≥2⁢a1+2⁢a2+4⁢a4+8⁢a8+⋯+2n⁢a2n\displaystyle\geq 2a_{1}+2a_{2}+4a_{4}+8a_{8}+\cdots+2^{n}a_{2^{n}}
=a1+(a1+2⁢a2+4⁢a4+8⁢a8+⋯+2n⁢a2n)\displaystyle=a_{1}+\left(a_{1}+2a_{2}+4a_{4}+8a_{8}+\cdots+2^{n}a_{2^{n}}\right)
=a1+tn+1.\displaystyle=a_{1}+t_{n+1}.

Replacing n+1n+1 with nn in the above and rearranging,

(3.7) (3.7) tn≤2⁢s2n−1−a1≤2⁢s2n−1for all n∈ℕ.t_{n}\leq 2s_{2^{n-1}}-a_{1}\leq 2s_{2^{n-1}}\qquad\text{for all $n\in\mathbb{% N}$.}

If ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges, then the sequence of partial sums (sn)n∈ℕ(s_{n})_{n\in\mathbb{N}} is bounded. By (3.7), the sequence of partial sums (tn)n∈ℕ(t_{n})_{n\in\mathbb{N}} is therefore also bounded. Consequently, ∑k=0∞2k⁢a2k\sum_{k=0}^{\infty}2^{k}a_{2^{k}} converges by the boundedness test from Lemma 3.26. ∎

Exercise 3.40.

Complete the proof of Theorem 3.38 by showing 2 ⇒\Rightarrow 1. Hint: this time, partition

s2n=(a1)⏟≤1⋅a1+(a2+a3)⏟≤2⋅a2+(a4+a6+a7+a7)⏟≤4⋅a4+⋯+(a2n−1+a2n−1+2+⋯+a2n−1)⏟≤2n−1⋅a2n−1+a2n.s_{2^{n}}=\underbrace{(a_{1})}_{\displaystyle\leq 1\cdot a_{1}}+\underbrace{(a% _{2}+a_{3})}_{\displaystyle\leq 2\cdot a_{2}}\\ +\underbrace{(a_{4}+a_{6}+a_{7}+a_{7})}_{\displaystyle\leq 4\cdot a_{4}}+% \cdots+\underbrace{(a_{2^{n-1}}+a_{2^{n-1}+2}+\cdots+a_{2^{n}}-1)}_{% \displaystyle\leq 2^{n-1}\cdot a_{2^{n-1}}}+a_{2^{n}}.
Exercise 3.41.

Determine whether the following series converge or diverge.

  1. (i)
    ​

    ∑k=2∞1k⁢(log2⁡k)2\displaystyle\sum_{k=2}^{\infty}\frac{1}{k(\log_{2}k)^{2}};

  2. (ii)
    ​

    ∑k=2∞1(log2⁡k)log2⁡k\displaystyle\sum_{k=2}^{\infty}\frac{1}{(\log_{2}k)^{\log_{2}k}}.

Corollary 3.42 (pp-series).

For p∈ℝp\in\mathbb{R}, the series ∑k=1∞1/kp\sum_{k=1}^{\infty}1/k^{p} converges if and only if p>1p>1.

Proof.

If p≤0p\leq 0 then the sequence of terms (1/kp)k∈ℕ(1/k^{p})_{k\in\mathbb{N}} is either unbounded (if p<0p<0) or all the terms are equal to 11 (if p=0p=0). In either case, the sequence does not converge to 0 and therefore the series ∑k=1∞1/kp\sum_{k=1}^{\infty}1/k^{p} diverges by the kkth term test from Lemma 3.18.

Now suppose p>0p>0. In this case, the sequence (1/kp)k∈ℕ(1/k^{p})_{k\in\mathbb{N}} is non-increasing and non-negative, so we may apply the condensation test. The condensed series is ∑k=0∞2k⋅1(2k)p{\sum_{k=0}^{\infty}2^{k}\cdot\frac{1}{(2^{k})^{p}}}. Observe that

2k⋅1(2k)p=1(2k)p−1=(2−(p−1))k.2^{k}\cdot\frac{1}{(2^{k})^{p}}=\frac{1}{(2^{k})^{p-1}}=(2^{-(p-1)})^{k}.

By Example 3.4, the geometric series ∑k=0∞(2−(p−1))k\sum_{k=0}^{\infty}(2^{-(p-1)})^{k} converges if and only if 2−(p−1)<12^{-(p-1)}<1, which is equivalent to p>1p>1. Thus, by the condensation test, ∑k=1∞1/kp\sum_{k=1}^{\infty}1/k^{p} converges if and only if p>1p>1. ∎

Example 3.43.

The series ∑k=1∞k+1−kk\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k} converges.

Proof.

We shall prove this by combining the comparison and pp-series tests. We use the bound on k+1−k\sqrt{k+1}-\sqrt{k} from Example 2.34 to obtain

(3.8) (3.8) 0<k+1−kk≤1k⋅1k=1k3/2for all k∈ℕ.0<\frac{\sqrt{k+1}-\sqrt{k}}{k}\leq\frac{1}{k}\cdot\frac{1}{\sqrt{k}}=\frac{1}% {k^{3/2}}\qquad\text{for all $k\in\mathbb{N}$.}

We apply the pp-series test with p=3/2>1p=3/2>1 to deduce that ∑k=1∞1/k3/2\sum_{k=1}^{\infty}1/k^{3/2} converges. Thus, ∑k=1∞k+1−kk\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k} converges by the comparison test.33 3 There are two perhaps more obvious comparisons, namely (3.9) (3.9) 0<k+1−kk≤1kand0<k+1−kk≤k+1−k.0<\frac{\sqrt{k+1}-\sqrt{k}}{k}\leq\frac{1}{k}\qquad\text{and}\qquad 0<\frac{% \sqrt{k+1}-\sqrt{k}}{k}\leq\sqrt{k+1}-\sqrt{k}. However, in the context of the comparison test, both lead to inconclusive results as the series formed from the terms on the right are divergent. Can you explain why, intuitively, (3.8) is more efficient than either of the inequalities in (3.9)? ∎

Exercise 3.44.

Determine whether the following series converge.

  1. (i)
    ​

    ∑k=1∞k1/3k2+10\displaystyle\sum_{k=1}^{\infty}\frac{k^{1/3}}{k^{2}+10};

  2. (ii)
    ​

    ∑k=1∞k2+1−kk1/100\displaystyle\sum_{k=1}^{\infty}\frac{\sqrt{k^{2}+1}-k}{k^{1/100}}.