7.3 Series

Tests

Theorem 217 Root test

[219] Let \(𝛼=\limsup _{n→∞}\sqrt[n]{|a_ n|}\) then

  • if \(𝛼{\lt}1\) the series \(∑_{n=1}^∞ a_ n\) converges absolutely;

  • if \(𝛼=1\) nothing can be concluded;

  • if \(𝛼{\gt}1\) the series \(∑_{n=1}^∞ a_ n\) does not converge, and also \(∑_{n=1}^∞ |a_ n|\) diverges.

Proof

[21B]

  • If \(𝛼{\lt}1\), having fixed \(L∈(𝛼,1)\) you have eventually \(\sqrt[n]{|a_{n}|}{\lt}L\) so there is a \(N\) for which \(|a_ n|≤ L^{N-n}\) for each \(n≥ N\) and we conclude by comparison with the geometric series.

  • For the two series \(1/n\) and \(1/n^ 2\) you have \(𝛼=1\).

  • If \(𝛼{\gt}1\) you have frequently \(\sqrt[n]{|a_{n}|}{\gt}1\) So \(|a_ n|{\gt}1\), contrary to the necessary criterion.

Theorem 218 Ratio test

[21C] Assume that \(a_ n\neq 0\). Let \(𝛼=\limsup _{n→∞}\frac{|a_{n+1}|}{|a_{n}|}\) then

  • if \(𝛼{\lt}1\) the series \(∑_{n=1}^∞ a_ n\) converges absolutely;

  • if \(𝛼≥ 1\) nothing can be concluded.

Proof
  • If \(𝛼{\lt}1\), taken \(L∈(𝛼,1)\) you have eventually \(\frac{|a_{n+1}|}{|a_{n}|}{\lt}L\) so there is a \(N\) for which \(\frac{|a_{n+1}|}{|a_{n}|}{\lt}L\) for each \(n≥ N\), by induction it is shown that \(|a_ n|≤ L^{n-N} |a_ N|\) and ends by comparison with the geometric series.

  • Let’s see some examples. For the two series \(1/n\) and \(1/n^ 2\) you have \(𝛼=1\).

    Defining

    \begin{equation} a_ n= \begin{cases} 2^{-n} & n~ \text{even}\\ 2^{2-n} & n~ \text{odd}\\ \end{cases} \label{eq:f3422p3sa} \end{equation}
    219

    we obtain a convergent series but for which \(𝛼=2\).

Remark 223

[0F1]If the ratio test 218 can be applied, we have seen in the demonstration that, for a \(L{\lt}1\), there is a \(N\) for which \(|a_ n|≤ L^{n-N}a_ N\) for every \(n≥ N\), and therefore \(\limsup _{n→∞}\sqrt[n]{|a_ n|}≤ L {\lt}1\), that is the root test 224 holds.

Theorem 224

[21D] If \((a_ n)_ n⊂{\mathbb {R}}\) has positive terms and is monotonic (weakly) decreasing, the series converges if and only if the series

\[ ∑_{n=1}^∞ 2^ n a_{2^ n} \]

converges.

Proof

Since the sequence \((a_ n)_ n\) is decreasing, then for \(h∈{\mathbb {N}}\)

\begin{equation} 2^{h}a_{2^{(h+1)}}≤ ∑_{k=2^ h+1}^{2^{(h+1)}}a_ k≤ 2^{h}a_{2^{h}}\quad .\label{eq:32rn2lp} \end{equation}
225

We note now that

\[ ∑_{h=0}^ N∑_{k=2^ h+1}^{2^{(h+1)}}a_ k = ∑_{n=2}^{2^{N+1}}a_ n \]

and therefore

\[ ∑_{h=0}^∞∑_{k=2^ h+1}^{2^{(h+1)}}a_ k= \lim _{N→∞} ∑_{h=0}^ N∑_{k=2^ h+1}^{2^{(h+1)}}a_ k= \lim _{N→∞} ∑_{n=2}^{2^{(N+1)}}a_ n {=} ∑_{n=2}^∞ a_ n \quad . \]

so we can add the terms in 225 to get

\[ ∑_{h=0}^∞ 2^{h}a_{2^{(h+1)}}≤ ∑_{n=2}^∞ a_ n≤∑_{h=0}^∞ 2^{h}a_{2^{h}} \]

where the term on the right is finite if and only if the one on the left is finite, because

\[ ∑_{h=0}^∞ 2^{h}a_{2^{h}}=a_ 1 + 2 ∑_{h=0}^∞ 2^{h}a_{2^{(h+1)}}\quad : \]

the proof ends by the comparison theorem

The Dirichlet criteria implies the Liebniz “alternating series test” criteria.

Theorem 226 Dirichlet criterion

[21F] Let \(\{ a_{n}\} \) and \(\{ b_{n}\} \) be two sequences. If \( b_{n}\) tends monotonically to \(0\) and if the series of partial sums of \(a_ n\) is bounded, i.e. if

\[ b_{n}≥ b_{n+1} {\gt}0\quad ,\quad \lim _{n→∞ } b_{n} = 0 \quad ,\quad ∃ M{\gt}0,~ ∀ N∈ℕ~ , \left|∑ _{n=1}^{N}a_{n}\right|{\lt}M\quad , \]

then the series

\[ ∑ _{n=1}^{+∞ }{a_{n}b_{n}} \]

is convergent.

The proof is left as an exercise (Hint: use 2)

Hidden solution: [UNACCESSIBLE UUID ’21G’]

In particular, if we set \(a_{n}=(-1)^{n}\) we prove the existence of the limit in Leibniz test.

Theorem 227 Alternating series test, or Leibniz test

[238] Let \(b_ n\) be a sequence for which

\[ b_{n}≥ b_{n+1} {\gt}0\quad , \quad \lim _{n→∞ } b_{n} = 0 \quad , \]

then the series

\[ ∑ _{n=0}^{+∞ }{ (-1)^{n}b_{n}} \]

is convergent; also, called \(ℓ\) the value of the series, letting

\[ B_ N = ∑ _{n=0}^{N }{ (-1)^{n}b_{n}} \]

the partial sums, we have that the sequence \(B_{2N}\) is decreasing , the sequence \(B_{2N+1}\) is increasing, and both converge to \(ℓ\).

Theorem 228

[0DR](Solved on 2022-12-13) Consider the series \(∑_{n=1}^∞ a_ n\) where the terms are positive: \(a_ n{\gt}0\). Define

\[ z_ n = n\left(\frac{a_ n}{a_{n+1}}-1\right) \]

for convenience.

  • If \(z_ n ≤ 1\) eventually in \(n\), then the series does not converge.

  • If there exists \(L{\gt}1\) such that \( z_ n≥ L\) eventually in \(n\), i.e. equivalently if

    \[ \liminf _{n→∞} z_ n{\gt}1\quad , \]

    then the series converges.

In addition, fixed \(h∈ {\mathbb {Z}}\), we can define

\[ z_ n = (n+h)\left(\frac{a_ n}{a_{n+1}}-1\right) \]

or

\[ z_ n = n\left(\frac{a_{n+h}}{a_{n+h+1}}-1\right) \]

such as

\[ z_ n = n\left(\frac{a_{n-1}}{a_{n}}-1\right) \]

and the criterion applies in the same way. Hidden solution: [UNACCESSIBLE UUID ’0DS’]

Exercises

E228

[214](Solved on 2022-12-13) Let \(𝛼{\gt}0\); use Raabe’s criterion 228 to study the convergence of the series

\[ ∑_{n=1}^∞ \frac{1}{n^𝛼} \]

Hidden solution: [UNACCESSIBLE UUID ’215’]

E228

[23D](Solved on 2022-12-13) Let \(𝛼{\gt}0\); use the condensation criterion 224 to study the convergence of the series

\[ ∑_{n=1}^∞ \frac{1}{n^𝛼} \]
E228

[0DW] Given a series \(∑_ n^∞ a_ n\) tell if the following conditions are necessary and/or sufficient for convergence.

\begin{eqnarray} ∀\varepsilon {\gt}0~ ∃ m∈{\mathbb {N}}~ ∀ n{\gt}m ~ ∀ k∈{\mathbb {N}}~ ~ \left|∑_{j=n}^{n+k} a_ k\right|{\lt}\varepsilon \\ ∀\varepsilon {\gt}0~ ∀ k∈{\mathbb {N}}~ ∃ m ∈{\mathbb {N}}~ ∀ n{\gt}m ~ \left|∑_{j=n}^{n+k} a_ k\right|{\lt}\varepsilon \\ ∀\varepsilon {\gt}0~ ∃ m∈{\mathbb {N}}~ ∀ n{\gt}m∀ k∈{\mathbb {N}}~ ~ ∑_{j=n}^{n+k} |a_ k| {\lt}\varepsilon \\ ∀\varepsilon {\gt}0~ ∀ k∈{\mathbb {N}}~ ∃ m∈{\mathbb {N}}~ ∀ n{\gt}m ~ ∑_{j=n}^{n+k} |a_ k|{\lt}\varepsilon \end{eqnarray}

Hidden solution: [UNACCESSIBLE UUID ’0DX’]

E228

[0DY](Proposed on 2022-12-13) Find two sequences \((a_ n)_ n,(b_ n)_ n\) with \(a_ n,b_ n{\gt}0\) such that \(∑_{n=0}^∞ (-1)^ n a_ n\) is convergent, \(∑_{n=0}^∞ (-1)^ n b_ n\) is non-convergent, and \(\lim _{n→∞} a_ n/b_ n=1\). Hidden solution: [UNACCESSIBLE UUID ’0DZ’]

E228

[0F0]Note:Exam of 9th APr 2011.Let \((a_ n)\) be a sequence of real numbers (not necessarily positive) such that the series \(∑_{n=1}^∞ a_ n\) converges to \(a∈{\mathbb {R}}\); let \(b_ n=\frac{a_ 1+\cdots +a_ n}{n}\); show that if the series \(∑_{n=1}^∞ b_ n\) converges then \(a=0\).

E228

[0F2](Proposed on 2022-12) Find two examples of \(a_{i,j}:{\mathbb {N}}× {\mathbb {N}}→ {\mathbb {R}}\)

  • such that, for each \(i\), \(∑_ j a_{i,j} =0\), while for each \(j\), \(∑_ i a_{i,j} =∞\);

  • such that, for each \(i\), \(∑_ j a_{i,j} =0\), while for each \(j\), \(∑_ i a_{i,j} =1\).

Can you find examples where moreover we have that \(|a_{i,j}|≤ 1\) for every \(i,j\)? [UNACCESSIBLE UUID ’0F3’]

[0F4]Note:Written exam of 4th Apr 2009, exee 1.(Proposed on 2022-12-13) Given a sequence \((a_ n)_{n}\) of strictly positive numbers, it is said that the infinite product \(∏_{n=0}^∞ a_ n\) converges if there exists finite and strictly positive the limit of partial products, i.e.

\[ \lim _{N→+∞}∏_{n=0}^ Na_ n ∈ (0,+∞)\quad . \]

Prove that

  1. if \(∏_{n=0}^∞ a_ n\) converges then \(\lim _{n→+∞}a_ n=1\);

  2. if the series \(∑_{n=0}^∞|a_ n-1|\) converges, then it also converges \(∏_{n=0}^∞ a_ n\);

  3. find an example where the series \(∑_{n=0}^∞(a_ n-1)\) converges but \(∏_{n=0}^∞ a_ n=0\).

[0F5]We indicate with \({\mathcal P}_{\mathfrak f}({\mathbb {N}})\) the set of subsets \(B⊆ {\mathbb {N}}\) which are finite sets. This is said the set of finite parts.

We abbreviate \({\mathcal P}={\mathcal P}_{\mathfrak f}({\mathbb {N}})\) in the following.

Given a sequence \((a_ n)_ n\) of real numbers and a \(B∈{\mathcal P}\) we indicate with \(s(B)=∑_{n∈ B} a_ n\) the finite sum with indices in \(B\).

Suppose the series \(∑_{n=0}^∞ a_ n\) converge but not converge at all. Then:

  • \(\{ s(F ) : F ∈{\mathcal P}\} \) it is dense in \({\mathbb {R}}\).

  • There is a reordering \(σ\) of \({\mathbb {N}}\), that is, a bijective function \(σ:{\mathbb {N}}→{\mathbb {N}}\), such that all partial sums \(∑_{n=0}^ N a_{σ(n)}\) (at the variation of \(N\)) is dense in \({\mathbb {R}}\).

[UNACCESSIBLE UUID ’0F6’] [0F7]Note:This result is attributed to Riemann , see 3.54 in [ 22 ] ..

Let be given a sequence \((a_ n)_ n\) of real numbers such that \(∑_{n=0}^∞ a_ n\) converges (to a finite value) but \(∑_{n=0}^∞ |a_ n|=∞\); for each \(l,L\) with \(-∞ ≤ l ≤ L ≤ +∞\) there is a permutation \(𝜋:{\mathbb {N}}→{\mathbb {N}}\) such that, defining \(S_ N=∑_{k=0}^ N a_{𝜋(k)}\), we have that

\[ \limsup _{N→∞} S_ N= L \quad , \quad \liminf _{N→∞} S_ N= l \quad . \]

[0F8]A sequence is given \((a_ n)_{n∈ {\mathbb {N}}}\) of positive real numbers such that \(\lim _{n→∞} a_ n=0\) and \(∑_{n=0}^∞ a_ n=∞\): prove that for every \(l ∈ {\mathbb {R}}\) there is a sequence \((ε_ n )_{n∈{\mathbb {N}}}\) with \(ε_ n ∈\{ 1,-1\} \) for each n, such that

\[ ∑_{n=0}^∞ (ε_ n a_ n)=l\quad . \]

If instead \(∑_{n=0}^∞ a_ n=S{\lt}∞\), what can be said about the set \(E\) of the sums \(∑_{n=0}^∞ (ε_ n a_ n)=l\), for all possible choices of \((ε_ n )_{n∈{\mathbb {N}}}\) with \(ε_ n ∈\{ 1,-1\} \) for every n?

  • Analyze cases where \(a_ n=2^{-n}\) or \(a_ n=3^{-n}\)

  • Show that \(E\) is always closed.

  • Under what assumptions do you have that \(E=[-S,S]\)?

Hint. Let \(\tilde E\) be the set of sums \(∑_ n (ε_ n a_ n)=l\), to vary by \((ε_ n )_{n∈{\mathbb {N}}}\) with \(ε_ n ∈ \{ 0,1\} \) for each n; note that \(\tilde E=\{ (S+x)/2 : x∈ E\} \). [0F9]Note:Written exam of 12th Jan 2019.

Show that the following series converges

\[ ∑_{n=1}^∞ \left(\frac{1⋅ 4⋅ 7⋅ 10 \cdots (3n-2)}{3⋅ 6⋅ 9 ⋅ 12 \cdots (3n)} \right)^ 2 \]

Hidden solution: [UNACCESSIBLE UUID ’0FB’] [21M](Proposed on 2022-12) Say for which \(𝛼{\gt}0,𝛽{\gt}0,𝛾{\gt}0\) you have that

\[ ∑_{n=4}^∞\frac 1{n^𝛼 \, (\log n)^𝛽\, (\log (\log n))^𝛾} \]

converges. [23F](Proposed on 2022-12-13) Note:Written exam 29th January 2021.Let it be \(𝛼{\gt}0\). Say (justifying) for which \(𝛼\) the following series converge or diverge

  • \[ ∑_{n=1}^∞ \left({\sqrt[4]{n^ 8+n^𝛼} - n^ 2 }\right) \]
  • \[ ∑_{n=2}^∞ \left( \frac{1}{n^𝛼} - \frac{1}{n^𝛼+1} \right) \]
  • \[ ∑_{n=2}^∞ \frac{1}{(\log _ 2 n) ^{𝛼\log _ 2(n)}} \]

    where the logarithms are in base 2.

Hidden solution: [UNACCESSIBLE UUID ’23G’] [20Z]Note:Task of 26 Jan 2016. (Solved on 2022-01-20)

Let

\[ z_ n = \frac{1⋅ 3⋅ 5⋅ 7 \cdots (2n-1)}{2⋅ 4⋅ 6⋅8 \cdots ~ (2n)} \quad ; \]

Show that \(\lim _{n→∞} z_ n=0\) but

\[ ∑_{n=1}^∞ z_ n=∞ \quad . \]

Hidden solution: [UNACCESSIBLE UUID ’213’] [210] Note:exercise 2, written exam 15 January 2014. Let \((a_ n)_{ n ≥ 0}\) be a sequence of positive real numbers. Having defined \(s_ n =∑_{i=0}^ n a_ i \) prove that:

  • the series \(∑_{n=0}^∞ a_ n\) converges if and only if the series \(∑_{n=0}^∞ a_{n}/s_ n\) converges;

  • the series \(∑_{n=0}^∞ a_ n / (s_ n)^ 2\) converges.

Hidden solution: [UNACCESSIBLE UUID ’21K’]

[UNACCESSIBLE UUID ’0DT’] [UNACCESSIBLE UUID ’0DV’] [UNACCESSIBLE UUID ’0FC’] [UNACCESSIBLE UUID ’0FD’] [UNACCESSIBLE UUID ’0FF’] [UNACCESSIBLE UUID ’0FG’] [UNACCESSIBLE UUID ’0DH’] See also exercise 1.

Cauchy product

Definition 233

[0FH]Give two sequences \((a_{n})_ n\) and \((b_{n})_ n\) to real or complex values, their Cauchy product is the sequence \((c_{n})_ n\) given by

\[ c_{n}{\stackrel{.}{=}}∑ _{k=0}^{n}a_{k}b_{n-k}\quad . \]

E233

[0FJ]If \(\forall n\in {\mathbb {N}}, a_ n,b_ n≥ 0\) show that

\[ ∑_{n=0}^∞ c_ n =∑_{n=0}^∞ a_ n ∑_{n=0}^∞ b_ n \]

with the convention that \(0⋅ ∞=0\).

E233

[0FK](Proposed on 2022-12-13) If the series \(∑_{n=0}^∞ a_ n\) and \(∑_{n=0}^∞ b_ n\) converge absolutely, show that the series \(∑_{n=0}^∞ c_ n\) converges absolutely and

\[ ∑_{n=0}^∞ c_ n =∑_{n=0}^∞ a_ n ∑_{n=0}^∞ b_ n\quad . \]
E233

[0FM] Prerequisites:1.Note:Known as: Mertens’ theorem..

If the series \(∑_{n=0}^∞ a_ n\) converges absolutely and \(∑_{n=0}^∞ b_ n\) converges, show that the series \(∑_{n=0}^∞ c_ n\) converges and

\[ ∑_{n=0}^∞ c_ n =∑_{n=0}^∞ a_ n ∑_{n=0}^∞ b_ n\quad . \]

Hidden solution: [UNACCESSIBLE UUID ’0FN’] [UNACCESSIBLE UUID ’218’]

[0FP] Discuss the Cauchy product of the series \(∑_{n=0}^∞ \frac{(-1)^{n}}{\sqrt{n+1}}\) with itself. Hidden solution: [UNACCESSIBLE UUID ’0FQ’]

See also exercise 1.