Infinite Series Convergence

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Convergence Tests for Infinite Series – HMC Calculus Tutorial

In this tutorial, we review some of the most common tests for the convergence of an infinite series k=0ak=a0+a1+a2+ The proofs or these tests are interesting, so we urge you to look them up in your calculus text.

Let s0=a0s1=a1sn=nk=0ak If the sequence {sn} of partial sums converges to a limit L, then the series is said to converge to the sum L and we write

k=0ak=L.

For j0, k=0ak converges if and only if k=jak converges, so in discussing convergence we often just write ak.

Example

Consider the geometric series k=0xk. The nth partial sum is sn=1+x+x2++xn. Multiplying both sides by x, xsn=x+x2+x3++xn+1. Subtracting the second equation from the first, (1x)sn=1xn+1, so for x1, sn=1xn+11x. For |x|<1, limnsn=11x. It is easy to see that k=0xk diverges for |x|1. Thus k=0xk=11x for |x|<1 and diverges for |x|1.

Divergence Test

If limkak0, then k=0ak diverges.

Example

The series k=0k2k+1 diverges, since limkk2k+1=1/20.

Integral Test

Let f(x) be continuous, decreasing, and positive for x1. Then k=1f(k) converges if and only if 1f(x)dx converges.

Example

Consider the p-series k=11kp=11p+12p+13p+ Since 11xpdx={11px1p|1p>1ln|x||1p=111px1p|10<p<1={11p, the series converges for p>1 and diverges for 0<p1.

The divergent p-series k=11k with p=1 is called the Harmonic Series.

Comparison Test

Let ak and bk be series with non-negative terms. If akbk for all k sufficiently large, then

  1. If bk converges, then ak also converges.
  2. If ak diverges, then bk also diverges.

Informally, if the “larger” series converges, so does the “smaller.” If the “smaller” series divers, so does the “larger.”

Examples
  • Since k=11k2 converges, so does k=11k2+3.
  • Since k=11k diverges, so does k=11ln|k+1|.

    1k2+3<1k2 for all k.

    1ln|k+1|>1k for k2.

    Limit Comparison Test

    Let ak and bk be series with positive terms. If limkakbk=L where 0<L< then ak and bk either both converge or both diverge.

    Example

    The series k=1k215k3 diverges, since k=11k diverges and limkk215k31k=limkk215k2=15.

    Ratio Test

    Let ak be a series with positive terms and suppose that limkak+1ak=L.

    1. If L<1, then ak converges.
    2. If L>1, then ak diverges.
    3. If L=1, then the test is inconclusive.

    Example

    The series k=11k! converges, since limk1(k+1)!1k!=limk1k+1=0.

    Root Test

    Let ak be a series with non-negative terms and suppose that limk(ak)1k=L.

    1. If L<1, then ak converges.
    2. If L>1, then ak diverges.
    3. If L=1, then the test is inconclusive.

    Example

    The series k=0(k2k+1)k converges, since limk[(k2k+1)k]1k=limkk2k+1=12.

    Alternating Series Test

    Consider the alternating series k=0(1)kak where ak>0 for all k0.

    If ak+1<ak for all k and limak=0, then k=0(1)kak converges.

    Example

    The series k=0(1)kk+1 converges, since 1(k+1)+1<1k+1 and limk1k+1=0. This series is conditionally convergent, rather than absolutely convergent, since k=0|(1)kk+1|=k=01k+1 diverges.


    Key Concepts

    The infinite series k=0ak converges if the sequence of partial sums converges and diverges otherwise.

    For a particular series, one or more of the common convergence tests may be most convenient to apply.


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