<Home>Teaching><MATH 31B (Fall 2020)>Week 9 Discussion Notes

Week 9 Discussion Notes

Table of Contents

Absolute Convergence

Definition

Let {an}\set{a_n} be a sequence.

  • If ∑n=0∞∣an∣\displaystyle \sum_{n=0}^\infty \abs{a_n} converges, then we say that the series is absolutely convergent.
  • If ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges, but it does not converge absolutely, then we say that the series is conditionally convergent.

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.

Integral Test

Theorem (integral test)

Let {an}\set{a_n} be a sequence and suppose that there is a function ff such that f(n)=anf\p{n} = a_n for all n≥Mn \geq M for some M>0M > 0. If f(x)f\p{x} is positive, decreasing, and continuous on [1,∞)\pco{1, \infty}, then

∑n=1∞an converges  ⟺  ∫1∞f(x) dx converges.\sum_{n=1}^\infty a_n \text{ converges} \iff \int_1^\infty f\p{x} \,\diff{x} \text{ converges}.

This one is pretty useful when it works, but as you can imagine, it can be really hard to use if your function is hard to integrate. The main application is convergence of pp-series:

Example 1.

Determine when ∑n=1∞1np\displaystyle \sum_{n=1}^\infty \frac{1}{n^p} converges.
Solution.

Here, f(x)=x−pf\p{x} = x^{-p}, which is continuous and non-negative when x≥1x \geq 1. Also, f′(x)=−px−p−1<0f'\p{x} = -px^{-p-1} < 0 if x≥1x \geq 1, too, so ff is decreasing on [1,∞)\pco{1, \infty}, which means that the integral test applies.

The improper integral ∫1∞1xp dx\int_1^\infty \frac{1}{x^p} \,\diff{x} converges if p>1p > 1 and diverges if p≤1p \leq 1 by the theorem on pp-integrals. By the integral test, this means that ∑n=1∞1np\sum_{n=1}^\infty \frac{1}{n^p} converges if and only if p>1p > 1.

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.

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, so by the limit comparison test, ∑n=1∞nn3+n2+7\sum_{n=1}^\infty \frac{n}{n^3 + n^2 + 7} converges also.

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)n−1bna_n = \p{-1}^{n-1}b_n, where bn>0b_n > 0 and bn≥bn+1b_n \geq b_{n+1} for n≥1n \geq 1. Then ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n converges.

In this case, ∑n=0∞an\displaystyle \sum_{n=0}^\infty a_n is called an alternating series, since the terms alternate signs at each step.

Example 3.

Determine whether ∑n=1∞(−1)nn\displaystyle \sum_{n=1}^\infty \frac{\p{-1}^n}{n} converges or not.
Solution.

We have an alternating series with bn=1nb_n = \frac{1}{n}, which decreases to 00. So, by the alternating series test, ∑n=1∞(−1)nn\sum_{n=1}^\infty \frac{\p{-1}^n}{n} converges.

This is an example of a conditionally convergent series, since the harmonic series diverges.

Midterm Review

Integration

Example 4.

Evaluate ∫ln⁡(x2+9) dx\displaystyle \int \ln\p{x^2 + 9} \,\diff{x}.
Solution.

Substitutions don't look very helpful here, so we're going to try integration by parts with u=ln⁡(x2+9)u = \ln\p{x^2 + 9} and dv=dx\diff{v} = \diff{x}. Then

du=2xx2+9 dxandv=x,\diff{u} = \frac{2x}{x^2 + 9} \,\diff{x} \quad\text{and}\quad v = x,

and the integral becomes

∫ln⁡(x2+9) dx=xln⁡(x2+9)−∫2x2x2+9 dx.\int \ln\p{x^2 + 9} \,\diff{x} = x\ln\p{x^2 + 9} - \int \frac{2x^2}{x^2 + 9} \,\diff{x}.

After doing a long division, we get

∫2x2x2+9 dx=∫2−18x2+9 dx=2x−6arctan⁡(x3)+C.\int \frac{2x^2}{x^2 + 9} \,\diff{x} = \int 2 - \frac{18}{x^2 + 9} \,\diff{x} = 2x - 6\arctan\p{\frac{x}{3}} + C.

You can integrate the fraction by using a table or using the trig sub x=3tan⁡θx = 3\tan\theta. Putting everything together,

∫ln⁡(x2+9) dx=xln⁡(x2+9)−2x+6arctan⁡(x3)+C.\int \ln\p{x^2 + 9} \,\diff{x} = \boxed{x\ln\p{x^2 + 9} - 2x + 6\arctan\p{\frac{x}{3}} + C}.
Example 5.

Evaluate ∫xx4+1 dx\displaystyle \int \frac{x}{x^4 + 1} \,\diff{x}.
Solution.

I see an xx in the numerator, so u=x2u = x^2 is probably a good start: du=2x dx\diff{u} = 2x \,\diff{x} and

∫xx4+1 dx=12∫duu2+1=12arctan⁡u+C=12arctan⁡(x2)+C.\int \frac{x}{x^4 + 1} \,\diff{x} = \frac{1}{2} \int \frac{\diff{u}}{u^2 + 1} = \frac{1}{2} \arctan{u} + C = \boxed{\frac{1}{2} \arctan\p{x^2} + C}.
Example 6.

Evaluate ∫cos⁡x2sin⁡2x+7sin⁡x+6 dx\displaystyle \int \frac{\cos{x}}{2\sin^2{x} + 7\sin{x} + 6} \,\diff{x}.
Solution.

Because of the cos⁡x\cos{x} in the numerator and all the sin⁡x\sin{x} terms in the denominator, I'm going to start with u=sin⁡x  ⟹  du=cos⁡x dxu = \sin{x} \implies \diff{u} = \cos{x} \,\diff{x}. The integral then turns into

∫cos⁡x2sin⁡2x+7sin⁡x+6 dx=∫du2u2+7u+6.\int \frac{\cos{x}}{2\sin^2{x} + 7\sin{x} + 6} \,\diff{x} = \int \frac{\diff{u}}{2u^2 + 7u + 6}.

You can factor 2u2+7u+6=(2u+3)(u+2)2u^2 + 7u + 6 = \p{2u + 3}\p{u + 2}, so by partial fractions, we end up with

12u2+7u+6=1(2u+3)(u+2)=22u+3−1u+2.\frac{1}{2u^2 + 7u + 6} = \frac{1}{\p{2u + 3}\p{u + 2}} = \frac{2}{2u + 3} - \frac{1}{u + 2}.

So, when we integrate, we just end up with

∫du2u2+7u+6=ln⁡(2u+3)−ln⁡(u+2)+C=ln⁡(2sin⁡x+3)−ln⁡(sin⁡x+2)+C.\begin{aligned} \int \frac{\diff{u}}{2u^2 + 7u + 6} &= \ln\p{2u + 3} - \ln\p{u + 2} + C \\ &= \boxed{\ln\p{2\sin{x} + 3} - \ln\p{\sin{x} + 2} + C}. \end{aligned}

Infinite Series

Example 7.

Calculate ∑n=−1∞2n+33n\displaystyle \sum_{n=-1}^\infty \frac{2^{n+3}}{3^n}.
Solution.

We can factor out a 232^3 to get

∑n=−1∞2n+33n=23∑n=−1∞2n3n=23∑n=−1∞(23)n=23(23)−11−23=8⋅32⋅3=36.\begin{aligned} \sum_{n=-1}^\infty \frac{2^{n+3}}{3^n} = 2^3 \sum_{n=-1}^\infty \frac{2^n}{3^n} &= 2^3 \sum_{n=-1}^\infty \p{\frac{2}{3}}^n \\ &= 2^3 \frac{\p{\frac{2}{3}}^{-1}}{1 - \frac{2}{3}} \\ &= 8 \cdot \frac{3}{2} \cdot 3 \\ &= \boxed{36}. \end{aligned}
Example 8.

Determine whether ∑n=0∞22nn!\displaystyle \sum_{n=0}^\infty \frac{2^{2n}}{n!} converges or not.
Solution.

As a rule of thumb, if you see factorials and/or powers of nn, the ratio test will usually work:

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

so by the ratio test, the series converges.
Example 9.

Determine whether ∑n=2∞1ln⁡3n\displaystyle \sum_{n=2}^\infty \frac{1}{\ln^3{n}} converges or not.
Solution.

The intuition here is that ln⁡n\ln{n} grows really, really slowly, which means that the terms here should go to 00 really, really slowly. So, this sum should diverge.

When you see logarithms, it's good to keep in mind that ln⁡n≤na\ln{n} \leq n^a for any a>0a > 0 if nn is large enough. If we cube both sides and rearrange the inequality, then

1n3a≤1ln⁡3n.\frac{1}{n^{3a}} \leq \frac{1}{\ln^3{n}}.

To use comparison, we want to pick aa so that the smaller integral diverges. a=13a = \frac{1}{3} works here: if nn is large enough, then

0≤1n≤1ln⁡3n,0 \leq \frac{1}{n} \leq \frac{1}{\ln^3{n}},

so by direct comparison, the series diverges.
Example 10.

Determine whether ∑n=1∞ln⁡nn2\displaystyle \sum_{n=1}^\infty \frac{\ln{n}}{n^2} converges or not.
Solution.

ln⁡n\ln{n} grows really slowly, so intuitively, it shouldn't affect the speed at which 1n2\frac{1}{n^2} goes to 00 by very much, so the series should converge.

The same idea works here: ln⁡n≤na\ln{n} \leq n^a if nn is large enough, so

ln⁡nn2≤nan2=1n2−a.\frac{\ln{n}}{n^2} \leq \frac{n^a}{n^2} = \frac{1}{n^{2-a}}.

To use comparison, we should pick aa so that the bigger series still converges. a=12a = \frac{1}{2} works here (we just need to make sure 2−a>12 - a > 1), so for nn large enough,

0≤ln⁡nn2≤1n3/2.0 \leq \frac{\ln{n}}{n^2} \leq \frac{1}{n^{3/2}}.

Thus, by direct comparison, the series converges.
Example 11.

Determine whether ∑n=1∞sin⁡(1n2)\displaystyle \sum_{n=1}^\infty \sin\p{\frac{1}{n^2}} converges or not.
Solution.

If you recall,

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

Intuitively, this means that if xx is really small, then sin⁡x≈x\sin{x} \approx x. If we replace xx with 1n2\frac{1}{n^2}, we get sin⁡(1n2)≈1n2\sin\p{\frac{1}{n^2}} \approx \frac{1}{n^2}, so the series should converge. When our intuition is based on approximations, the limit test will usually work:

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

I used the limit above to calculate this: as long as whatever's inside sin⁡x\sin{x} and the denominator are the same and goes to 00, the limit is 11.

When nn is large enough, 1n2<π\frac{1}{n^2} < \pi, so sin⁡(1n2)≥0\sin\p{\frac{1}{n^2}} \geq 0 when this happens. Thus, by limit comparison, the series converges.