Numerical Series Notes

CH1: Numerical Series

Objectives

  • Calculate sums in simple or classic cases.
  • Study the nature of a numerical series using different criteria:
    • Series with positive terms.
    • Riemann's criterion.
    • d'Alembert's rule.
    • Cauchy's rule.

Introductory Activity: Fourier Series

  • A particular form of numerical series used to decompose a periodic function into an infinite sum of trigonometric terms.
  • Essential in various fields such as signal processing and data compression.
  • Data compression is an essential technique in signal processing and modern communication systems.
  • It reduces the amount of data needed to represent a signal while retaining essential information.
  • Compressed signals include images, videos, and sounds.
  • Compression:
    • Decreases the memory space required to store data.
    • Accelerates data processing and transfer while maintaining high quality.
Question
  • Why are compressed audio or video files (MP3, MPEG) smaller than the originals while remaining very clear?
Answer
  • Any complex signal (e.g., a sound wave or an image) can be represented as a sum of simple sinusoidal waves, thanks to Fourier series.
Example
  • Let the signal be a square wave (e.g., a binary 0/1 signal) periodic with period T=2πT = 2\pi.
    S(t)={1if 0≤t≤T/2 0if T/2≤t≤TS(t) = \begin{cases} 1 & \text{if } 0 \leq t \leq T/2 \ 0 & \text{if } T/2 \leq t \leq T \end{cases}
  • Can be decomposed into a Fourier series (numerical series):
    S(t)=4π∑n=1∞1nsin⁡(nt)S(t) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin(nt), with n is odd.
  • In compression, only the most important sinusoids are retained, which reduces the amount of data.
  • The convergence of Fourier series guarantees that the infinite sum is well-defined and can be used to reconstruct a signal.

Generalities on Numerical Series

Definition
  • Let n<em>0∈Nn<em>0 \in \mathbb{N} and (U</em>n)<em>n≥n</em>0(U</em>n)<em>{n \geq n</em>0} be a real sequence.
    • For n≥n<em>0n \geq n<em>0, S</em>n=∑<em>k=n</em>0nU<em>kS</em>n = \sum<em>{k=n</em>0}^{n} U<em>k is called the partial sum of (U</em>n)<em>n≥n</em>0(U</em>n)<em>{n \geq n</em>0} of order n.
    • The sequence of partial sums (S<em>n)</em>n≥n<em>0(S<em>n)</em>{n \geq n<em>0}, denoted ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n, is called the numerical series of general term UnU_n.
Example 1
  1. The partial sum of the series with general term U<em>n=1n2U<em>n = \frac{1}{n^2} for n≥1n \geq 1 is defined by: S</em>n=∑k=1n1k2S</em>n = \sum_{k=1}^{n} \frac{1}{k^2}
  2. The partial sum of the series with general term U<em>n=1(n−2)(n−1)(n−3)U<em>n = \frac{1}{\sqrt{(n - 2)(n - 1)(n - 3)}} for n≥4n \geq 4 is defined by: S</em>n=∑k=4n1(k−2)(k−1)(k−3)S</em>n = \sum_{k=4}^{n} \frac{1}{\sqrt{(k - 2)(k - 1)(k - 3)}}
Convergence
  • Determining the nature of a series means determining whether it is convergent or divergent.
  • The convergence of the general term (U<em>n)</em>n≥n<em>0(U<em>n)</em>{n \geq n<em>0} does not generally imply the convergence of the numerical series ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n.
  • In particular, studying the convergence of ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n consists of studying the convergence of (S</em>n)<em>n≥n</em>0(S</em>n)<em>{n \geq n</em>0} and not of (U<em>n)</em>n≥n0(U<em>n)</em>{n \geq n_0}.
Example 2
  • Express S<em>nS<em>n as a function of n, then calculate lim⁡</em>n→+∞Sn\lim</em>{n \to +\infty} S_n.

    1. S<em>n=∑</em>k=1n1k(k+1)S<em>n = \sum</em>{k=1}^{n} \frac{1}{k(k + 1)}
    2. S<em>n=∑</em>k=0n12kS<em>n = \sum</em>{k=0}^{n} \frac{1}{2^k}
    3. S<em>n=∑</em>k=1nln⁡(1+1k)S<em>n = \sum</em>{k=1}^{n} \ln(1 + \frac{1}{k})
    4. S<em>n=∑</em>k=1ne−kS<em>n = \sum</em>{k=1}^{n} e^{-k}
  • If lim⁡<em>n→+∞S</em>n\lim<em>{n \to +\infty} S</em>n exists and is finite, we say that the sequence (Sn)(S_n) is convergent.

  • If lim⁡<em>n→+∞S</em>n\lim<em>{n \to +\infty} S</em>n is infinite or does not exist, we say that the sequence (Sn)(S_n) is divergent.

  • The series ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n converges to S if the sequence of partial sums S</em>n=∑<em>k=n</em>0nU<em>kS</em>n = \sum<em>{k=n</em>0}^{n} U<em>k converges to S. In this case, the limit S is called the sum of the series, and we write ∑</em>k=n<em>0∞U</em>k=lim⁡<em>n→∞S</em>n=S\sum</em>{k=n<em>0}^{\infty} U</em>k = \lim<em>{n \to \infty} S</em>n = S

  • The notations ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n and ∑</em>n=n<em>0+∞U</em>n\sum</em>{n=n<em>0}^{+\infty} U</em>n designate two different notions:

    • The notation ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n designates a numerical series.
    • In the case where it converges, the notation ∑<em>n=n</em>0+∞Un\sum<em>{n=n</em>0}^{+\infty} U_n designates its limit.
Remarque 4 (Attention!)
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a numerical series converging to S=∑</em>n=n<em>0+∞U</em>nS = \sum</em>{n=n<em>0}^{+\infty} U</em>n. We call the remainder of order n of this series the real number R<em>nR<em>n defined by: R</em>n=S−S<em>n=∑</em>k=n<em>0+∞U</em>k−∑<em>k=n</em>0nU<em>k=∑</em>k=n+1+∞UkR</em>n = S - S<em>n = \sum</em>{k=n<em>0}^{+\infty} U</em>k - \sum<em>{k=n</em>0}^{n} U<em>k = \sum</em>{k=n+1}^{+\infty} U_k
Exercice 1
  • Let ∑<em>n≥0U</em>n\sum<em>{n \geq 0} U</em>n be a numerical series converging to S=∑<em>n=n</em>0+∞UnS = \sum<em>{n=n</em>0}^{+\infty} U_n.

    1. For all n≥1n \geq 1, calculate R<em>n−1−R</em>nR<em>{n-1} - R</em>n.
    2. Calculate the limit of RnR_n as n tends to infinity.
    3. Conclude.
Necessary Condition for Convergence
  • To study the convergence of a numerical series ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n, it is necessary that its general term U</em>nU</em>n satisfies a certain condition that we define in what follows.
Activity 1
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a convergent numerical series, and S</em>n=∑<em>k=n</em>0nUkS</em>n = \sum<em>{k=n</em>0}^{n} U_k its partial sum of order n.

    1. Verify that U<em>n=S</em>n−Sn−1U<em>n = S</em>n - S_{n-1}.
    2. Calculate lim⁡<em>n→+∞U</em>n\lim<em>{n \to +\infty} U</em>n and conclude.
  • If the series ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n is convergent then (U</em>n)<em>n≥n</em>0(U</em>n)<em>{n \geq n</em>0} is convergent and lim⁡<em>n→+∞U</em>n=0\lim<em>{n \to +\infty} U</em>n = 0.

Exercice 2
  • Show that the following series are divergent.

    1. ∑n≥0n2−1n2+1\sum_{n \geq 0} \frac{n^2 - 1}{n^2 + 1}
    2. ∑n≥0arctan⁡(n)\sum_{n \geq 0} \arctan(n)
    3. ∑n≥0cos⁡(nπ)\sum_{n \geq 0} \cos(n\pi)
  • A numerical series whose general term does not tend towards 0 is divergent.

  • If lim⁡<em>n→+∞U</em>n=0\lim<em>{n \to +\infty} U</em>n = 0, is the series ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n convergent?

Exercice 3
  • For all n≥1n \geq 1, we set U<em>n=ln⁡(1+1n)U<em>n = \ln(1 + \frac{1}{n}) and S</em>n=∑<em>k=1nU</em>kS</em>n = \sum<em>{k=1}^{n} U</em>k.

    1. Calculate lim⁡<em>n→+∞U</em>n\lim<em>{n \to +\infty} U</em>n.
    2. Verify that ln⁡(1+1k)=ln⁡(k+1)−ln⁡(k)\ln(1 + \frac{1}{k}) = \ln(k + 1) - \ln(k), for k≥1k \geq 1.
    3. Calculate SnS_n.
    4. Calculate lim⁡<em>n→+∞S</em>n\lim<em>{n \to +\infty} S</em>n and deduce the nature of the series ∑<em>n≥1U</em>n\sum<em>{n \geq 1} U</em>n.
    5. What can we conclude?

Algebraic Operations on Series

  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n and ∑</em>n≥n<em>0V</em>n\sum</em>{n \geq n<em>0} V</em>n be two numerical series, and λ∈R\lambda \in \mathbb{R}.
∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n∑<em>n≥n</em>0Vn\sum<em>{n \geq n</em>0} V_n∑<em>n≥n</em>0(U<em>n+λV</em>n)\sum<em>{n \geq n</em>0} (U<em>n + \lambda V</em>n)
convergesconvergesconverges
convergesdivergesdiverges
divergesconvergesdiverges
divergesdivergesWe cannot conclude
Example 3
  • Study the nature of each of the following numerical series:

    1. ∑n≥2(1n(n−1)+12n)\sum_{n \geq 2} \left(\frac{1}{n(n - 1)} + \frac{1}{2n}\right)
    2. ∑n≥2(1n(n−1)+arctan⁡(n))\sum_{n \geq 2} \left(\frac{1}{n(n - 1)} + \arctan(n)\right)
    3. ∑n≥2(2n+1n−2n−1n−1)\sum_{n \geq 2} \left(\frac{2n + 1}{n} - \frac{2n - 1}{n - 1}\right)

Reference Series

  • In this section, we will present some reference series.
Geometric Series
Activity 2
  • Let q∈Rq \in \mathbb{R}, n<em>0∈Nn<em>0 \in \mathbb{N} and S</em>n=∑<em>k=n</em>0nqkS</em>n = \sum<em>{k=n</em>0}^{n} q^k.

    1. Calculate S<em>nS<em>n for all n≥n</em>0n \geq n</em>0.
    2. Study, according to the values of q, the limit of SnS_n as n tends towards +∞+\infty.
  • The geometric series ∑<em>n≥n</em>0qn\sum<em>{n \geq n</em>0} q^n is convergent if and only if q∈]−1,1[q \in ]-1, 1[ and in this case, ∑<em>n=n</em>0+∞qn=qn01−q\sum<em>{n=n</em>0}^{+\infty} q^n = \frac{q^{n_0}}{1 - q}.

Example 4
  • Study the nature of each of the following series:

    1. ∑n≥0(910)n\sum_{n \geq 0} \left(\frac{9}{10}\right)^n
    2. ∑n≥05n\sum_{n \geq 0} 5^n
    3. ∑n≥0(−13)n\sum_{n \geq 0} \left(\frac{-1}{3}\right)^n
Telescopic Series
  • Let (a<em>n)</em>n≥n<em>0(a<em>n)</em>{n \geq n<em>0} be a sequence. The series with general term (a</em>n+1−an)(a</em>{n+1} - a_n) is called a telescopic series
Activity 3
  • Let n<em>0∈Nn<em>0 \in \mathbb{N} and (a</em>n)<em>n≥n</em>0(a</em>n)<em>{n \geq n</em>0} be a real sequence.

    1. Express ∑<em>k=n</em>0n(a<em>k+1−a</em>k)\sum<em>{k=n</em>0}^{n} (a<em>{k+1} - a</em>k) as a function of a<em>n+1a<em>{n+1} and a</em>n0a</em>{n_0}.
    2. Show then that the series ∑<em>n≥n</em>0(a<em>n+1−a</em>n)\sum<em>{n \geq n</em>0} (a<em>{n+1} - a</em>n) and the sequence (a<em>n)</em>n≥n0(a<em>n)</em>{n \geq n_0} have the same nature.
  • The telescopic series ∑<em>n≥n</em>0(a<em>n+1−a</em>n)\sum<em>{n \geq n</em>0} (a<em>{n+1} - a</em>n) and the sequence (a<em>n)</em>n≥n0(a<em>n)</em>{n \geq n_0} are of the same nature.

Exercice 4
  • Study the nature of each of the following series:

    1. ∑n≥21(n−1)n\sum_{n \geq 2} \frac{1}{(n - 1)n}
    2. ∑n≥1ln⁡(1−1n+1)\sum_{n \geq 1} \ln(1 - \frac{1}{n + 1})
Riemann Series
  • For α∈R\alpha \in \mathbb{R} and n<em>0∈N∗n<em>0 \in \mathbb{N}^*, a Riemann series is a series defined by ∑</em>n≥n01nα\sum</em>{n \geq n_0} \frac{1}{n^\alpha}.
Activity 4
  • We seek to determine a convergence criterion for Riemann series.

    1. Suppose that α≤0\alpha \leq 0, calculate lim⁡<em>n→+∞1nα\lim<em>{n \to +\infty} \frac{1}{n^\alpha} and deduce that the series ∑</em>n≥11nα\sum</em>{n \geq 1} \frac{1}{n^\alpha} is divergent.
    2. Suppose that α>0\alpha > 0. Let k≥1k \geq 1 be an integer, noting that the function t↦1tαt \mapsto \frac{1}{t^\alpha} is decreasing on [k,k+1][k, k + 1], show that
      1(k+1)α≤1tα≤1kα∀t∈[k,k+1]\frac{1}{(k + 1)^\alpha} \leq \frac{1}{t^\alpha} \leq \frac{1}{k^\alpha} \quad \forall t \in [k, k + 1]
    3. By integrating the preceding inequality between k and k + 1, then summing from k = 1 to k = n, show that
      ∑<em>k=2n+11kα≤∫</em>1n+1dttα≤∑k=1n1kα\sum<em>{k=2}^{n+1} \frac{1}{k^\alpha} \leq \int</em>{1}^{n+1} \frac{dt}{t^\alpha} \leq \sum_{k=1}^{n} \frac{1}{k^\alpha}
    4. Calculate ∫1n+1dttα\int_{1}^{n+1} \frac{dt}{t^\alpha} as a function of α\alpha.
    5. Discuss, according to the values of α\alpha, the nature of the Riemann series ∑n≥11nα\sum_{n \geq 1} \frac{1}{n^\alpha}.
  • The Riemann series ∑n≥11nα\sum_{n \geq 1} \frac{1}{n^\alpha} is convergent if and only if α>1\alpha > 1.

Exercice 5
  • Study the nature of the following numerical series:

    1. ∑n≥11n\sum_{n \geq 1} \frac{1}{\sqrt{n}}
    2. ∑n≥11nn3\sum_{n \geq 1} \frac{1}{n\sqrt{n^3}}
Bertrand Series
  • A Bertrand series is a series defined by ∑n≥21nα⋅(ln⁡(n))β\sum_{n \geq 2} \frac{1}{n^\alpha \cdot (\ln(n))^\beta} where (α,β)∈R2(\alpha, \beta) \in \mathbb{R}^2.
Activity 5 (an idea of proof)
  1. Suppose that α<1\alpha < 1. We have lim⁡n→+∞n×1nα⋅(ln⁡(n))β=+∞\lim_{n \to +\infty} n \times \frac{1}{n^\alpha \cdot (\ln(n))^\beta} = +\infty, show that the Bertrand series is divergent.

  2. Suppose that α>1\alpha > 1. We have lim⁡n→+∞nr×1nα⋅(ln⁡(n))β=0\lim_{n \to +\infty} n^r \times \frac{1}{n^\alpha \cdot (\ln(n))^\beta} = 0 where 1<r<α1 < r < \alpha, show that the Bertrand series is convergent.

  3. Suppose that α=1,β≤0\alpha = 1, \beta \leq 0. Show that for all n≥3n \geq 3 we have: 1n≤1n⋅(ln⁡(n))β\frac{1}{n} \leq \frac{1}{n \cdot (\ln(n))^\beta}, and deduce that the Bertrand series is divergent.

  4. Suppose that α=1\alpha = 1 and β>1\beta > 1.

    a) Show that for all n≥3n \geq 3 we have: 1n⋅(ln⁡(n))β≤∫n−1ndtt⋅(ln⁡(t))β\frac{1}{n \cdot (\ln(n))^\beta} \leq \int_{n-1}^{n} \frac{dt}{t \cdot (\ln(t))^\beta}.

    b) Deduce that for all N≥3N \geq 3 we have: ∑n=3N1n⋅(ln⁡(n))β≤1β−1⋅1(ln⁡(2))β−1\sum_{n=3}^{N} \frac{1}{n \cdot (\ln(n))^\beta} \leq \frac{1}{\beta - 1} \cdot \frac{1}{(\ln(2))^{\beta-1}}, then conclude that the Bertrand series converges.

  • Let (α,β)∈R2(\alpha, \beta) \in \mathbb{R}^2. The Bertrand series ∑n≥21nα⋅(ln⁡(n))β\sum_{n \geq 2} \frac{1}{n^\alpha \cdot (\ln(n))^\beta} converges if and only if:

  • α>1\alpha > 1, or (α=1\alpha = 1 and β>1\beta > 1).

Exercice 6
  • Give the nature of the following numerical series:

    1. ∑n≥21n⋅ln⁡(n)\sum_{n \geq 2} \frac{1}{\sqrt{n} \cdot \ln(n)}
    2. ∑n≥2ln⁡(n)n2\sum_{n \geq 2} \frac{\ln(n)}{n^2}
    3. ∑n≥21n⋅(ln⁡(n))2\sum_{n \geq 2} \frac{1}{n \cdot (\ln(n))^2}
    4. ∑n≥2ln⁡(n)n\sum_{n \geq 2} \frac{\ln(n)}{n}

Series with Positive Terms

  • In this section, we group together some convergence criteria that are only valid for series with positive terms.
  • A real series ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n is said to have positive terms if U</em>n≥0,∀n∈NU</em>n \geq 0, \quad \forall n \in \mathbb{N}.
Activity 6
  • Let n<em>0∈Nn<em>0 \in \mathbb{N}, ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n be a series with positive terms, and S<em>n=∑</em>k=n<em>0nU</em>kS<em>n = \sum</em>{k=n<em>0}^{n} U</em>k its partial sum of order n.

    1. Calculate S<em>n+1−S</em>nS<em>{n+1} - S</em>n, for all n≥n0n \geq n_0.
    2. Show that the sequence (Sn)(S_n) is increasing.
    3. Conclude.
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a series with positive terms, then, ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n converges if and only if the sequence of partial sums (Sn)(S_n) is bounded above.

Exercice 7
  • Let the sequence U<em>n=∫</em>n−1n11+t2dt,∀n∈N∗U<em>n = \int</em>{n-1}^{n} \frac{1}{1 + t^2} dt, \quad \forall n \in \mathbb{N}^*.

    1. Show that Un≥0,∀n∈N∗U_n \geq 0, \quad \forall n \in \mathbb{N}^*.
    2. Calculate S<em>n=∑</em>k=1nUkS<em>n = \sum</em>{k=1}^{n} U_k.
    3. Show that the sequence (Sn)(S_n) is bounded above.
    4. Deduce the nature of the series ∑<em>n≥1U</em>n\sum<em>{n \geq 1} U</em>n.
Comparison Criteria
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n and ∑</em>n≥n<em>0V</em>n\sum</em>{n \geq n<em>0} V</em>n be two series with positive terms such that:
    0≤U<em>n≤V</em>n,∀n∈N0 \leq U<em>n \leq V</em>n , \forall n \in \mathbb{N}

  • If ∑<em>n≥n</em>0V<em>n\sum<em>{n \geq n</em>0} V<em>n converges then ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n converges.

  • If ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n diverges then ∑</em>n≥n<em>0V</em>n\sum</em>{n \geq n<em>0} V</em>n diverges.

Exercice 8
  • Study the nature of each of the following series:

    1. ∑n≥1sin⁡2(n)n2\sum_{n \geq 1} \frac{\sin^2(n)}{n^2}
    2. ∑n≥11ncos⁡2(n)\sum_{n \geq 1} \frac{1}{n \cos^2(n)}
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n and ∑</em>n≥n<em>1V</em>n\sum</em>{n \geq n<em>1} V</em>n be two series with positive terms such that U<em>nU<em>n and V</em>nV</em>n are equivalent in the neighborhood of +∞+\infty:
    U<em>n∼</em>+∞V<em>n  ⟺  lim⁡</em>n→+∞U<em>nV</em>n=1U<em>n \sim</em>{+ \infty} V<em>n \iff \lim</em>{n \to +\infty} \frac{U<em>n}{V</em>n} = 1

  • Then, ∑<em>n≥n</em>1V<em>n\sum<em>{n \geq n</em>1} V<em>n and ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n are of the same nature.

Exercice 9
  • Study the nature of the following series:

    1. ∑n≥1n2+1n5+7n2+8\sum_{n \geq 1} \frac{n^2 + 1}{n^5 + 7n^2 + 8}
    2. ∑n≥1ln⁡(1+1n2)\sum_{n \geq 1} \ln\left(1 + \frac{1}{n^2}\right)
    3. ∑n≥11n(n2+1)\sum_{n \geq 1} \frac{1}{n(n^2 + 1)}
Riemann's Criterion
Activity 7
  • Let ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n be a series with positive terms and α∈R\alpha \in \mathbb{R}.

    1. Suppose that α≤1\alpha \leq 1 and lim⁡<em>n→+∞nαU</em>n=+∞\lim<em>{n \to +\infty} n^\alpha U</em>n = +\infty.

      a) Show that there exists N<em>1≥n</em>0N<em>1 \geq n</em>0, such that U<em>n≥1nα,∀n≥N</em>1U<em>n \geq \frac{1}{n^\alpha} , \forall n \geq N</em>1.

      b) Deduce the nature of the series ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n.

    2. Suppose that α>1\alpha > 1 and lim⁡<em>n→+∞nαU</em>n=0\lim<em>{n \to +\infty} n^\alpha U</em>n = 0.

      a) Show that there exists N<em>2≥n</em>0N<em>2 \geq n</em>0 such that 0≤U<em>n≤1nα,∀n≥N</em>20 \leq U<em>n \leq \frac{1}{n^\alpha} , \forall n \geq N</em>2.

      b) Deduce the nature of the series ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n.

  • Let ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n be a series with positive terms.

    • If there exists α≤1\alpha \leq 1 such that lim⁡<em>n→+∞nαU</em>n≠0\lim<em>{n \to +\infty} n^\alpha U</em>n \neq 0 then the series ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n is divergent.
    • If there exists α>1\alpha > 1 such that lim⁡<em>n→+∞nαU</em>n≠∞\lim<em>{n \to +\infty} n^\alpha U</em>n \not= \infty (exists and is finite) then the series ∑<em>n≥0U</em>n\sum<em>{n \geq 0} U</em>n is convergent
Exercice 10
  • Study the nature of each of the following series:

    1. ∑n≥0ne−n\sum_{n \geq 0} ne^{-n}
    2. ∑n≥1ln⁡(1+1n)\sum_{n \geq 1} \ln \left(1 + \frac{1}{n}\right)
    3. ∑n≥1ln⁡(1+1n2)\sum_{n \geq 1} \ln \left(1 + \frac{1}{n^2}\right)
d'Alembert's Rules
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a series with strictly positive terms such that lim⁡</em>n→+∞U<em>n+1U</em>n=l\lim</em>{n \to +\infty} \frac{U<em>{n+1}}{U</em>n} = l

    • If l<1l < 1 then ∑<em>n≥0U</em>n\sum<em>{n \geq 0} U</em>n is convergent.
    • If l>1l > 1 then ∑<em>n≥0U</em>n\sum<em>{n \geq 0} U</em>n is divergent.
    • If l=1l = 1, we cannot conclude.
Example 5 (Case lim⁡<em>n→+∞U</em>n+1Un=1\lim<em>{n \to +\infty} \frac{U</em>{n+1}}{U_n} = 1)
  • Let U<em>n=1nαU<em>n = \frac{1}{n^\alpha}, we have lim⁡</em>n→+∞U<em>n+1U</em>n=1\lim</em>{n \to +\infty} \frac{U<em>{n+1}}{U</em>n} = 1 but ∑n≥11nα\sum_{n \geq 1} \frac{1}{n^\alpha} converges if α>1\alpha > 1 and diverges if α≤1\alpha \leq 1.
Exercice 11
  • Study the nature of each of the following series:

    1. ∑n≥1n!nn\sum_{n \geq 1} \frac{n!}{n^n}
    2. ∑n≥2nan,a∈R∗\sum_{n \geq 2} na^n, a \in \mathbb{R}^*
Cauchy's Rules
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a series with strictly positive terms such that lim⁡</em>n→+∞Unn=l\lim</em>{n \to +\infty} \sqrt[n]{U_n} = l

    • If l<1l < 1 then ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n is convergent.
    • If l>1l > 1 then ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n is divergent.
    • If l=1l = 1, we cannot conclude.
Example 6 (Case lim⁡<em>n→+∞U</em>nn=1\lim<em>{n \to +\infty} \sqrt[n]{U</em>n} = 1)
  • Let U<em>n=1nαU<em>n = \frac{1}{n^\alpha}, we have lim⁡</em>n→+∞U<em>nn=1\lim</em>{n \to +\infty} \sqrt[n]{U<em>n} = 1 but ∑</em>n≥11nα\sum</em>{n \geq 1} \frac{1}{n^\alpha} converges if α>1\alpha > 1 and diverges if α≤1\alpha \leq 1.
Exercice 12
  • Study the nature of each of the following series:

    1. ∑n≥12n−13n+1\sum_{n \geq 1} \frac{2^{n-1}}{3^{n+1}}
    2. ∑n≥21(ln⁡(n))n\sum_{n \geq 2} \frac{1}{(\ln(n))^n}
  • Series with Negative Terms

  • A real series ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n is said to have negative terms if U</em>n≤0,∀n≥n0U</em>n \leq 0 , \forall n \geq n_0.

  • For the study of the convergence of this series, we can restrict ourselves to the case of the study of the convergence of a series with positive terms by setting V<em>n=−U</em>nV<em>n = -U</em>n.

  • Thus, all the convergence criteria used for the study of the nature of a real positive series are valid in the case where the real series has terms of constant sign.

Series with Arbitrary Terms

  • When the real series ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n is not with terms of constant sign, the criteria stated previously are no longer valid.
  • In this case, we need other criteria to study its convergence.
Alternating Series
  • A series is said to be alternating if it is of one of the following types:
    ∑<em>n≥n</em>0(−1)nU<em>n or ∑</em>n≥n<em>0(−1)n+1U</em>n\sum<em>{n \geq n</em>0} (-1)^n U<em>n \text{ or } \sum</em>{n \geq n<em>0} (-1)^{n+1} U</em>n
    where U<em>n≥0∀n≥n</em>0U<em>n \geq 0 \quad \forall n \geq n</em>0
Example 7
  • The following series are alternating series:
    ∑<em>n≥1(−1)nn,∑</em>n≥1cos⁡(nπ)n\sum<em>{n \geq 1} \frac{(-1)^n}{\sqrt{n}} , \sum</em>{n \geq 1} \frac{\cos(n\pi)}{n}.

  • Let ∑<em>n≥n</em>0(−1)nU<em>n\sum<em>{n \geq n</em>0} (-1)^n U<em>n be an alternating series, if the sequence (U</em>n)<em>n≥n</em>0(U</em>n)<em>{n \geq n</em>0} is decreasing and lim⁡<em>n→+∞U</em>n=0\lim<em>{n \to +\infty} U</em>n = 0, then, ∑<em>n≥n</em>0(−1)nUn\sum<em>{n \geq n</em>0} (-1)^n U_n is convergent.

  • Moreover, we have ∣R<em>n∣=∣∑</em>k=n+1+∞(−1)nU<em>k∣≤U</em>n+1,∀n≥n0\vert R<em>n\vert = \vert \sum</em>{k=n+1}^{+\infty} (-1)^n U<em>k\vert \leq U</em>{n+1} , \forall n \geq n_0

Exercice 13
  • Study the nature of each of the following series:

    1. ∑n≥1(−1)nn\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}
    2. ∑n≥1(−1)nn2\sum_{n \geq 1} \frac{(-1)^n}{n^2}
    3. ∑n≥1cos⁡(nπ)e−n\sum_{n \geq 1} \cos(n\pi)e^{-n}
Absolute Convergence
  • To bring the study of a series with arbitrary terms back to that of a series with positive terms, we define the absolute convergence of a numerical series.
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a numerical series with arbitrary terms, we say that the series ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n is absolutely convergent if ∑<em>n≥n</em>0∣Un∣\sum<em>{n \geq n</em>0} \vert U_n \vert is convergent.
Exercice 14
  • Show that the series ∑n≥1cos⁡(n)n2\sum_{n \geq 1} \frac{\cos(n)}{n^2} is absolutely convergent.

  • If ∑<em>n≥n</em>0Un\sum<em>{n \geq n</em>0} U_n is an absolutely convergent series then it is convergent.

Exercice 15
  • Let the series ∑n≥1(−1)nn\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}.

    1. Show that the series ∑n≥1(−1)nn\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}} is convergent.
    2. Study the absolute convergence of the series ∑n≥1(−1)nn\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}.
    3. Conclude.
  • The converse of the theorem is false, there exist series that are convergent and not absolutely convergent.

Semi-convergence
  • Let ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n be a series with arbitrary terms, we say that the series ∑</em>n≥n<em>0U</em>n\sum</em>{n \geq n<em>0} U</em>n is semi-convergent if ∑<em>n≥n</em>0U<em>n\sum<em>{n \geq n</em>0} U<em>n is convergent and ∑</em>n≥n<em>0∣U</em>n∣\sum</em>{n \geq n<em>0} \vert U</em>n \vert is divergent.
Application
  • A person suffering from an illness must take a daily dose of 20mg of a certain medication. Each day, the body eliminates 25% of the medication present.

  • We denote qnq_n the quantity of medication (in mg) present in the body at the end of the nth day (n∈N∗n \in \mathbb{N}^*).

    1. Show that, for all n∈N∗n \in \mathbb{N}^*, q<em>n=20∑</em>k=1n(34)kq<em>n = 20 \sum</em>{k=1}^{n} (\frac{3}{4})^k.
    2. The body can only support a quantity of 70mg of this medication. Knowing that she must take this medication for the rest of her days, is this person in danger?