Divergence Test Study Notes

Divergence Test Overview

  • The test for Divergence, also known as the Divergence Test, is a method used to determine the divergence of a series quickly.

  • It is particularly useful as it saves time by immediately identifying that a series is divergent without requiring rigorous convergence proofs.

Statement of the Divergence Test

  • The Divergence Test states:

    • If the limit of the sequence of terms a<em>na<em>n does not exist, or if it exists but is not equal to zero, then the series a</em>n\sum a</em>n is a divergent series.

    • Consequently, the only scenario under which a series could converge is if its terms converge to zero.

Proof of the Divergence Test

  • To prove this test, we will demonstrate an equivalent statement:

    • If a series is convergent, then the limit of its terms must be zero.

Understanding Equivalence

  • It’s critical to recognize why these statements are equivalent:

    • Divergence Test: If the limit of ana_n is not zero (either it doesn’t exist or equals a non-zero number), then the series diverges.

    • Convergence Implication: If the series converges, it must mean that the terms approach zero; otherwise, the series cannot converge.

Setup for Proof

  • Assume we have a convergent series represented by the summation of terms ana_n.

  • Let SnS_n represent the nth partial sum of the series, defined as:

    • S<em>n=a</em>1+a<em>2++a</em>nS<em>n = a</em>1 + a<em>2 + … + a</em>n

  • The nth term can be expressed as:

    • a<em>n=S</em>nSn1a<em>n = S</em>n - S_{n-1}

Convergence of Partial Sums

  • Since the series converges, the sequence of partial sums SnS_n is also convergent.

  • By definition:

    • lim<em>nS</em>n=S\lim<em>{n \to \infty} S</em>n = S

  • Also:

    • lim<em>nS</em>n1=S\lim<em>{n \to \infty} S</em>{n-1} = S

  • This can be established since S<em>n1S<em>{n-1} is just one term prior to S</em>nS</em>n, and as n approaches infinity, both limits approach the same value S.

Finding the Limit of Sequence Terms

  • Given the equation for ana_n, we can find:

    • lim<em>na</em>n=lim<em>n(S</em>nSn1)\lim<em>{n \to \infty} a</em>n = \lim<em>{n \to \infty} (S</em>n - S_{n-1})

  • By applying the limit properties, this becomes:

    • lim<em>na</em>n=lim<em>nS</em>nlim<em>nS</em>n1\lim<em>{n \to \infty} a</em>n = \lim<em>{n \to \infty} S</em>n - \lim<em>{n \to \infty} S</em>{n-1}

  • Substituting the limits we established earlier yields:

    • lim<em>na</em>n=SS=0\lim<em>{n \to \infty} a</em>n = S - S = 0

Conclusion of Proof

  • We have shown that if a series converges, then the limit of its terms must equal zero:

    • lim<em>na</em>n=0\lim<em>{n \to \infty} a</em>n = 0

  • Therefore, according to the Divergence Test:

    • If the limit of the terms does not equal zero, the series must be divergent.

Important Note on Converse

  • It is vital to note that the converse of this statement is not true:

    • If we know only that the limit of the terms approaches zero, this does not imply that the series converges.

    • Example: The limit of 1n\frac{1}{n} as nn approaches infinity is zero, however, the sum of all these terms (the harmonic series) diverges.

Summary of Divergence Test Applications

  • If the limit of a sequence approaches zero, we cannot definitively conclude whether the series converges or diverges; the series may do either.

  • However, if the limit of the terms does not approach zero, then we can assert that the series is divergent.