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Week 5 Discussion Notes

Table of Contents

Convergence Tests

For your convenience, I'm going to restate all of the convergence tests here.

Divergence Test

Theorem (divergence test)

Let {an}\set{a_n} be a sequence. If lim⁡n→∞an\displaystyle \lim_{n\to\infty} a_n does not exist or if it exists and is not 00, then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n diverges.

This should always be the first thing you try. Most of the time, it'll be inconclusive, though.

Direct Comparison Test

Theorem (direct comparison test)

Suppose 0≤an≤bn0 \leq a_n \leq b_n for n≥Mn \geq M for some M>0M > 0. Then:

  • If ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n diverges, then ∑n=0∞bn\displaystyle \sum_{n=0}^\infty b_n diverges as well.
  • If ∑n=0∞bn\displaystyle \sum_{n=0}^\infty b_n converges, then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges as well.

Limit Comparison Test

Theorem (limit comparison test)

Suppose an≥0a_n \geq 0 and bn≥0b_n \geq 0 for n≥Mn \geq M for some M>0M > 0. Let L=lim⁡n→∞anbn\displaystyle L = \lim_{n\to\infty} \frac{a_n}{b_n}.

  • If L∈(0,∞)L \in \p{0, \infty}, then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges if and only if ∑n=0∞bn\displaystyle \sum_{n=0}^\infty b_n converges.
  • If L=0L = 0 and ∑n=0∞bn\displaystyle \sum_{n=0}^\infty b_n converges, then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges as well.
  • If L=∞L = \infty and ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges, then ∑n=0∞bn\displaystyle \sum_{n=0}^\infty b_n converges as well.

The case when L=(0,∞)L = \p{0, \infty} is when this is most useful, in my experience. If L=0L = 0 or L=∞L = \infty, you can usually use direct comparison instead.

Ratio Test

Theorem (ratio test)

Let L=lim⁡n→∞∣an+1an∣\displaystyle L = \lim_{n\to\infty} \abs{\frac{a_{n+1}}{a_n}}.

  • If L<1L < 1, then ∑n=1∞an\displaystyle \sum_{n=1}^\infty a_n converges absolutely.
  • If L>1L > 1, then ∑n=1∞an\displaystyle \sum_{n=1}^\infty a_n diverges.
  • If L=1L = 1, then the ratio test is inconclusive.

Root Test

Theorem (root test)

Let L=lim⁡n→∞∣an∣1/n\displaystyle L = \lim_{n\to\infty} \abs{a_n}^{1/n}.

  • If L<1L < 1, then ∑n=1∞an\displaystyle \sum_{n=1}^\infty a_n converges absolutely.
  • If L>1L > 1, then ∑n=1∞an\displaystyle \sum_{n=1}^\infty a_n diverges.
  • If L=1L = 1, then the root test is inconclusive.

The ratio test and root test are pretty similar in that if one is inconclusive, the other one probably will be, too. They also "measure" how fast the terms ana_n converge to 00.

  • If L<1L < 1, then the terms converge fast enough that the whole series converges.

  • If L>1L > 1, then the terms converge too slow so the series diverges.

  • If L=1L = 1, the rate is somewhere in between too fast and too slow, which is why they're inconclusive.

Alternating Series Test

Theorem (alternating series test)

Let {an}\set{a_n} be a sequence of the form an=(−1)nbna_n = \p{-1}^n b_n, where bn≥0b_n \geq 0, is decreasing, and lim⁡n→∞bn=0\displaystyle\lim_{n\to\infty} b_n = 0. Then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges.

Examples

The best way to really learn these tests is to practice using them.

Example 1.

Determine whether ∑n=1∞12n+3n\displaystyle \sum_{n=1}^\infty \frac{1}{2^n + 3^n} converges.
Solution.

If I remove 3n3^n from the denominator, I get something bigger:

12n+3n≤12n.\frac{1}{2^n + 3^n} \leq \frac{1}{2^n}.

All the terms are non-negative, and ∑n=1∞12n\sum_{n=1}^\infty \frac{1}{2^n} converges because it's a geometric series with ∣r∣=12<1\abs{r} = \frac{1}{2} < 1. So, by direct comparison, I get that ∑n=1∞12n+3n\sum_{n=1}^\infty \frac{1}{2^n + 3^n} also converges.
Example 2.

Determine whether ∑n=1∞nn3+n2+7\displaystyle \sum_{n=1}^\infty \frac{n}{n^3 + n^2 + 7} converges.
Solution.

Intuitively, when nn is very large, the main terms that matter are the leading terms:

nn3+n2+7≈nn3=1n2,\frac{n}{n^3 + n^2 + 7} \approx \frac{n}{n^3} = \frac{1}{n^2},

so intuitively, this should converge. The limit test makes this rigorous:

lim⁡n→∞nn3+n2+71n2=lim⁡n→∞n3n3+n2+7⋅1n31n3=lim⁡n→∞11+1n+7n3=1.\begin{aligned} \lim_{n\to\infty} \frac{\frac{n}{n^3 + n^2 + 7}}{\frac{1}{n^2}} &= \lim_{n\to\infty} \frac{n^3}{n^3 + n^2 + 7} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\ &= \lim_{n\to\infty} \frac{1}{1 + \frac{1}{n} + \frac{7}{n^3}} \\ &= 1. \end{aligned}

∑n=1∞1n2\sum_{n=1}^\infty \frac{1}{n^2} converges because it's a pp-series with p=2>1p = 2 > 1, so by the limit comparison test, ∑n=1∞nn3+n2+7\sum_{n=1}^\infty \frac{n}{n^3 + n^2 + 7} converges also.
Example 3.

Determine whether ∑n=0∞22nn!\displaystyle \sum_{n=0}^\infty \frac{2^{2n}}{n!} converges.
Solution.

As a rule of thumb, if you ever see exponentials and/or factorials, the root or ratio test will do the trick. In this problem, the limit in the ratio test is easier to calculate:

lim⁡n→∞∣an+1an∣=lim⁡n→∞22(n+1)(n+1)!⋅n!22n=lim⁡n→∞22n+222n⋅n!(n+1)!=lim⁡n→∞22n+1=0.\begin{aligned} \lim_{n\to\infty} \abs{\frac{a_{n+1}}{a_n}} &= \lim_{n\to\infty} \frac{2^{2\p{n+1}}}{\p{n+1}!} \cdot \frac{n!}{2^{2n}} \\ &= \lim_{n\to\infty} \frac{2^{2n+2}}{2^{2n}} \cdot \frac{n!}{\p{n+1}!} \\ &= \lim_{n\to\infty} \frac{2^2}{n+1} \\ &= 0. \end{aligned}

00 is smaller than 11, so the ratio test tells us that our series converges.
Example 4.

Determine whether ∑n=1∞(−1)n1n\displaystyle \sum_{n=1}^\infty \p{-1}^n \frac{1}{\sqrt{n}} converges.
Solution.

Whenever you see something like (−1)n\p{-1}^n, chances are, you will want to use the alternating series test. To use it, there're just three things to check:

  1. 1n≥0\frac{1}{\sqrt{n}} \geq 0
  2. 1n\frac{1}{\sqrt{n}} is decreasing
  3. lim⁡n→∞1n=0\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0

These are all clear because of the properties of n\sqrt{n}, so by the alternating series test, the series converges.
Example 5.

Determine whether ∑n=1∞(−1)nn22n2−1\displaystyle \sum_{n=1}^\infty \p{-1}^n \frac{n^2}{2n^2 - 1} converges.
Solution.

It might be tempting to immediately try the alternating series test, but it won't work for this problem. You should always start with the divergence test: in this problem,

lim⁡n→∞(−1)n⏟divergesn22n2−1⏟converges to 12=DNE.\lim_{n\to\infty} \underbrace{\p{-1}^n}_{\text{diverges}} \underbrace{\frac{n^2}{2n^2 - 1}}_{\text{converges to } \frac{1}{2}} = \dne.

So, the divergence test tells us that the series diverges.
Example 6.

Determine whether ∑n=1∞sin⁡(1n2)\displaystyle \sum_{n=1}^\infty \sin\p{\frac{1}{n^2}} converges.
Solution.

To solve this, you'll want to remember this limit:

lim⁡x→0sin⁡xx=1.\lim_{x\to0} \frac{\sin{x}}{x} = 1.

Heuristically, this tells you that if x≈0x \approx 0, then

sin⁡xx≈1  ⟹  sin⁡x≈x.\frac{\sin{x}}{x} \approx 1 \implies \sin{x} \approx x.

In our problem, if nn is really large, then 1n2≈0\frac{1}{n^2} \approx 0, so

sin⁡(1n2)≈1n2.\sin\p{\frac{1}{n^2}} \approx \frac{1}{n^2}.

This tells us what to use with limit comparison:

lim⁡n→∞∣sin⁡(1n2)1n2∣=1,\lim_{n\to\infty} \abs{\frac{\sin\p{\frac{1}{n^2}}}{\frac{1}{n^2}}} = 1,

and because ∑n=1∞1n2\sum_{n=1}^\infty \frac{1}{n^2} is a pp-series with p=2>1p = 2 > 1, it converges. Thus, by the limit comparison test, ∑n=1∞sin⁡(1n2)\sum_{n=1}^\infty \sin\p{\frac{1}{n^2}} converges (absolutely).
Example 7.

True or false:

If ∑nan\displaystyle \sum_n a_n and ∑nbn\displaystyle \sum_n b_n both converge, then ∑nanbn\displaystyle \sum_n a_nb_n converges.
Solution.

This is false. For example, if an=bn=(−1)n1na_n = b_n = \p{-1}^n \frac{1}{\sqrt{n}}, then both ∑nan,∑nbn\sum_n a_n, \sum_n b_n converge by Example 4, but

∑nanbn=∑n1n\sum_n a_nb_n = \sum_n \frac{1}{n}

is the harmonic series, which diverges.