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=a1⋮sn=n∑k=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 j≥0, ∞∑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, (1−x)sn=1−xn+1, so for x≠1, sn=1−xn+11−x. For |x|<1, limn→∞sn=11−x. It is easy to see that ∞∑k=0xk diverges for |x|≥1. Thus ∞∑k=0xk=11−x for |x|<1 and diverges for |x|≥1.
Divergence Test
If limk→∞ak≠0, then ∞∑k=0ak diverges.
Example
The series ∞∑k=0k2k+1 diverges, since limk→∞k2k+1=1/2≠0.
Integral Test
Let f(x) be continuous, decreasing, and positive for x≥1. 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={11−px1−p|∞1p>1ln|x||∞1p=111−px1−p|∞10<p<1={11−p∞∞, the series converges for p>1 and diverges for 0<p≤1.
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 ak≤bk for all k sufficiently large, then
- If ∑bk converges, then ∑ak also converges.
- 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
1k2+3<1k2 for all k.
1ln|k+1|>1k for k≥2.
Limit Comparison Test
Let ∑ak and ∑bk be series with positive terms. If limk→∞akbk=L where 0<L<∞ then ∑ak and ∑bk either both converge or both diverge.
Example
The series ∞∑k=1k2−15k3 diverges, since ∞∑k=11k diverges and limk→∞k2−15k31k=limk→∞k2−15k2=15.
Ratio Test
Let ∑ak be a series with positive terms and suppose that limk→∞ak+1ak=L.
- If L<1, then ∑ak converges.
- If L>1, then ∑ak diverges.
- If L=1, then the test is inconclusive.
Example
The series ∞∑k=11k! converges, since limk→∞1(k+1)!1k!=limk→∞1k+1=0.
Root Test
Let ∑ak be a series with non-negative terms and suppose that limk→∞(ak)1k=L.
- If L<1, then ∑ak converges.
- If L>1, then ∑ak diverges.
- If L=1, then the test is inconclusive.
Example
The series ∞∑k=0(k2k+1)k converges, since limk→∞[(k2k+1)k]1k=limk→∞k2k+1=12.
Alternating Series Test
Consider the alternating series ∞∑k=0(−1)kak where ak>0 for all k≥0.
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 limk→∞1k+1=0. This series is conditionally convergent, rather than absolutely convergent, since ∞∑k=0|(−1)kk+1|=∞∑k=01k+1 diverges.
Key Concepts
For a particular series, one or more of the common convergence tests
may be most convenient to apply.
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