$$$$

Assignment 8

Exercise 1 For which $x\in \mathbb{R}$ does the following series converge? $$1+x^2+x^4+x^8+x^{16}+x^{32}+x^{64}+\cdots =\sum _{k\ge 0}{x}^{2^k}$$

Proof. We can prove divergence immediately for $|x|\ge 1$, by noting $|x|\ge 1$ implies $|x{|}^{2^k}\ge 1$ and so $|x{|}^{2^k}$ cannot converge to zero. This implies $x^{2^k}$ does not converge to zero (the contrapositive of $|a_k|\to 0⟹a_k\to 0$), so $\sum x^{2^k}$ diverges.

For $|x|<1$, we apply the ratio test, which compares with a geometric series. $$\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{x^{2^{k+1}}}{x^{2^k}}\right|=\left|\frac{x^{2^k\cdot 2}}{x^{2^k}}\right|=\left|\frac{{\left(x^{2^k}\right)}^2}{x^{2^k}}\right|=\left|x^{2^k}\right|=|x{|}^{2^k}$$ When $|x|<1$ this sequence is a subsequence of $|x{|}^n$ (for $n$ being powers of 2), and $|x{|}^n\to 0$ is one of our basic limits. Since all subsequences of a convergent sequence converge to the same limit, we see $|x{|}^{2^k}\to 0$, and $0<1$ implies our series converges.

Thus our series converges on $(-1,1)$ and diverges at every point outside this interval, so the domain is $(-1,1)$.

Exercise 2 Define the sequence $d_n$ to run through the digits $1-9$ and then repeat periodically, so $$d_n=1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,\dots$$ Find the radius of convergence of the power series $$\sum _{k\ge 1}{d}_k{x}^k$$ (Hint: can you find a series which is always $\ge$ and another which is always $\le$, and do some comparison and convergence tests?

There are a variety of proof techniques one could consider on this and related problems with series. Here’s two example solutions:

Proof (Comparison). To show the radius of convergence is 1 we just need to show it converges for |x|<1$ and diverges for $|x|>1$. In both cases it will be helpful to consider series of absolute values $\sum _k|d_k{x}^k|=\sum _k{d}_k|x{|}^k$.

• First, when $|x|>1$ then $|x{|}^k>1^k=1$. As $d_k\ge 1$ this implies $d_k|x{|}^k>1$ for all $k$ so $d_k|x{|}^k$ cannot converge to zero. Hence $d_k{x}^k$ also does not converge to zero, so our series diverges by the divergence test.

• Secondly, when $|x|<1$, we use comparison. Since $d_k\le 9$ we see $d_k|x{|}^k<9|x{|}^k$, and so $\sum d_k|x{|}^k\le \sum 9|x{|}^k=9\sum |x{|}^k$. Thus our series is bounded above by a multiple of the geometric series, which converges on $(-1,1)$, and by comparison we see our series must also converge on this interval. Thus $\sum _k{d}_k{x}^k$ converges absolutely on $(-1,1)$ and hence converges there.

Note One could also use the comparison test for the inequality $\sum |x{|}^k\le \sum d_k|x{|}^k$ to prove divergence for $|x|>1$ of the absolute value of the series, but would have to be more careful to ensure it works for the series itself, when $x$ is negative.

Proof (Cauchy Hadamard). One may also apply the general theorem for power series, which we know (always works) at the expense of having to compute a limsup. For this problem, that implies computing $lim sup\sqrt[k]{|d_k|}$. Since for all $k$ we know that $1\le d_k\le 9$ we have $\sqrt[k]{1}\le \sqrt[k]{d_k}\le \sqrt[k]{9}$. Then, using the inequality of limits (if $a_n\le b_n$ then $lim sup{a}_n\le lim sup{b}_n$), we see $$lim sup\sqrt[k]{1}\le lim sup\sqrt[k]{d_k}\le lim sup\sqrt[k]{9}$$

The lower bound here is just the constant sequence $1$, as $\sqrt[k]{1}=1$ for all $k$. And the upper bound is also 1, by our basic $a^{1/k}\to 1$ for any positive $a$. Thus, the limsup must also equal 1, and our radius of convergence is $$R=\frac{1}{lim sup\sqrt[k]{d_k}}=\frac{1}{1}=1$$

Exercise 3 This problem shows that for a power series, the domain may have any sort of behavior at the endpoints (open, closed, half open, half closed, etc).

In class we saw $\sum _{k\ge 0}{x}^k$ converges on the interval $(-1,1)$. Show that

• $\sum _{k\ge 1}\frac{x^k}{k}$ converges on $[-1,1)$
• $\sum _{k\ge 1}\frac{x^k}{k^2}$ converges on $[-1,1]$
• Construct an example of a power series which converges on $(-1,1]$ (and prove your claim).

Proof.

1. Consider the series $\sum _{k\ge 1}\frac{x^k}{k}$, Using the ratio test $$|\frac{x^{k+1}}{k+1}\cdot \frac{k}{x^k}|=|x|\cdot \frac{k}{k+1}\to |x|\text{as }k\to \mathrm{\infty }.$$ So, the series converges absolutely when $|x|<1$, and diverges when $|x|>1$. To analyze the endpoints:

• At $x=1$, we have the harmonic series $\sum \frac{1}{k}$, which diverges.
• At $x=-1$, we have the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$, which converges by the alternating series test (the terms decrease monotonically to zero in absolute value).

Now we have checked every real number, and confiremd the series converges on $[-1,1)$.

2. Consider the series $\sum _{k\ge 1}\frac{x^k}{k^2}.$ By the ratio test, $$|\frac{x^{k+1}}{(k+1{)}^2}\cdot \frac{k^2}{x^k}|=|x|\cdot {\left(\frac{k}{k+1}\right)}^2\to |x|\text{as }k\to \mathrm{\infty }.$$ So, the series converges absolutely for $|x|<1$, and diverges for $|x|>1$.

• At $x=1$, we have $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ which converges (a result proven in class).
• At $x=-1$, we have $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k^2},$ which also converges absolutely (since the sum of $1/k^2$ converges, the alternating version does as well).

Thus, the series converges on $[-1,1]$.

3. Define the power series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^k}{k}.$ Which is just the series from (a) where now each term is multiplied by $(-1{)}^n$. Since the ratio test involves absolute values, it will give exactly the same answers as part (a), showing convergence for $|x|<1$ and divergence for $|x|>1$. But now the endpoint behavior has reversed:

• At $x=1$ the series becomes the alternating harmonic series, which converges by the alternating series test.
• At $x=-1$ the series becomes the harmonic series, which diverges.

Hence, the series converges on $(-1,1]$, as required.

Exercise 4 In class we proved that for a sequence $a_n\in (-1,1)$ with $lim{a}_n=a\in (-1,1)$ $$\underset{n}{lim}\sum _{k\ge 0}(a_n{)}^k=\sum _{k\ge 0}\underset{n}{lim}(a_n{)}^k$$ By using the fact that we know a closed formula for the geometric sum $\sum _{k\ge 0}{x}^k=\frac{1}{1-x}$.

In this problem, re-prove the same fact but using Dominated Convergence

*Hint: if $a_n$ and $a$ are in $(-1,1)$ first prove that for some fixed $r<1$ that $a_n$ is eventually less than $r$ for all larger $n$. Then set $M_k=r^k$ and confirm all the hypotheses of dominated convergence.

Proof. Because $a_n\to a$ with $a\in (-1,1)$, for every $\epsilon >0$ there exists $N$ such that for all $n\ge N$, $|a_n-a|<\epsilon .$ Choose $\epsilon =\frac{1-|a|}{2}$, and set $r=\frac{1+|a|}{2}$. Since $a\in (-1,1)$, we have $\epsilon >0$ and $r<1$. Hence, for some $N$ we assure $|a_n-a|<ϵ$, which after some algebra implies $|a_n|<r$ for all $n>N$.

Set $M_k=r^k$. Then for all $n$, we have $|(a_n{)}^k|\le r^k=M_k$.

The series $ {k } M_k = {k } r^k$ converges since $r<1$ is within the radius of convergence.

Thus we’ve provided the dominating series required to invoke Dominated Convergence. Applying it, we switch the limit and the sum to give $$\underset{n\to \mathrm{\infty }}{lim}\sum _{k\ge 0}(a_n{)}^k=\sum _{k\ge 0}\underset{n\to \mathrm{\infty }}{lim}(a_n{)}^k.$$

As desired.

Exercise 5 (Cauchy’s Double Summation Formula) Use Dominated Convergence to prove the double summation formula: If $a_{m,n}$ be a double sequence, and either $$\sum _{m\ge 0}\sum _{n\ge 0}|a_{m,n}|\text{or}\sum _{n\ge 0}\sum _{m\ge 0}|a_{m,n}|$$ converges. Then both iterated sums exist and are equal: $$\sum _{m\ge 0}\sum _{n\ge 0}{a}_{m,n}=\sum _{n\ge 0}\sum _{m\ge 0}{a}_{m,n}$$

Hint: without loss of generality, assume that $\sum _{m\ge 0}\sum _{n\ge 0}|a_{m,n}|$ converges. Set $M_m=\sum _{n\ge 0}|a_{m,n}|$ and show the various hypotheses of Dominated convergence apply

Our assumption is then that $\sum _{m\ge 0}\sum _{n\ge 0}|a_{m,n}|$ converges, and we wish to show (1) that $\sum _{m\ge 0}\sum _{n\ge 0}{a}_{m,n}$ converges, and (2) so does the iterated sum, after we exchange the order of summation. We do each part separately for clarity.

Solution (Convergence of $\sum _m\sum _n{a}_{m,n}$). Let’s introduce some helpful notation: let $s_m=\sum _{n\ge 0}{a}_{m,n}$, and $M_m=\sum _{n\ge 0}|a_{m,n}|$ the sum of the absolute values. As absolute convergence implies convergence, it is enough to prove that $\sum _{m\ge 0}{s}_m$ converges absolutely; or $\sum _{m\ge 0}|s_m|$ converges. This is a sum of nonnegative terms, so we can use comparison. For each $m$, we use the triangle inequality on partial sums and the limit inequalities to see $$|s_m|=\left|\sum _{n\ge 0}{a}_{m,n}\right|\le \sum _{n\ge 0}|a_{m,n}|=M_m$$

As our assumption was precisely that $\sum _m{M}_m$ converges, comparison tells us $\sum _m|s_m|$ does as well, so $\sum _m{s}_m$ is absolutely convergent, and hence convergent. Unpacking our notation once more, $\sum _m\sum _n{a}_{m,n}$ converges, as required.

Solution (Showing the Two Orders are Equal). Now, on to (2). Dominated convergence lets us switch the order of an infinte sum and a limit, so to be able to apply the theorem, we need to rewrite one of our infinite sums using its definition, as a limit of finite sums.

$$\sum _{m\ge 0}\sum _{n\ge 0}{a}_{m,n}=\underset{K}{lim}\sum _{m=0}^K\sum _{n\ge 0}{a}_{m,n}$$

For any fixed value of $K$, we then have a finite sum of infinite series

$$\sum _{m=0}^K\sum _{n\ge 0}{a}_{m,n}=\sum _{n\ge 0}{a}_{0,n}+\sum _{n\ge 0}{a}_{1,n}+\cdots +\sum _{n\ge 0}{a}_{K,n}$$

As each of these sums converges, we can collect them all into a single sum (by the limit law for sums):

$$=\sum _{n\ge 0}(a_{0,n}+a_{1,n}+\cdots +a_{K,n})=\sum _{n\ge 0}\sum _{m=0}^K{a}_{m,n}$$

Stringing these inequalities together makes it clearer that dominated convergence will be useful:

$$\sum _{m\ge 0}\sum _{n\ge 0}{a}_{m,n}=\underset{K}{lim}\sum _{m=0}^K\sum _{n\ge 0}{a}_{m,n}=\underset{K}{lim}\sum _{n\ge 0}\sum _{m=0}^K{a}_{m,n}$$

Assume for a second that we can switch the order of limit and sum. Then this would equal

$$\sum _{n\ge 0}\underset{K}{lim}\sum _{m=0}^K{a}_{m,n}=\sum _{n\ge 0}\sum _{m\ge 0}{a}_{m,n}$$

So this really is all we need to justify. Seeing this is the bulk of the problem, as to justify using dominated convergence is immediate using the hint. Looking back at the theorem statement, we need to check that the terms $\sum _{m=0}^K{a}_{m,n}$ being summed up are uniformly bounded by some $M_n$ such that $\sum _n{M}_n$ converges. Following the hint, we take $M_n=\sum _{m\ge 0}|a_{m,n}|$. Then

$$\left|\sum _{m=0}^K{a}_{m,n}\right|\le \sum _{m=0}^K|a_{m,n}|\le \sum _{m\ge 0}|a_{m,n}|=M_n$$ Where the last inequality follows as we are summing nonnegative terms, so the adding more only increases the value. Thus, $|\sum _{m=0}^K{a}_{m,n}|<M_n$ and $\sum _n{M}_n$ converges, so dominated convergence allows the limit switch we need!

Exercise 6 (Abel’s Test) Prove that if the series $\sum _{k=1}^{\mathrm{\infty }}{x}_k$ converges, and if $(y_k)$ is a monotone decreasing nonnegative sequence, then the series $\sum _{k=1}^{\mathrm{\infty }}{x}_k{y}_k$ converges.

Hint: Use Abel’s Lemma, which states $\sum _{k=m+1}^n{x}_k{y}_k=s_n{y}_n-s_m{y}_m-\sum _{k=m}^{n-1}{s}_k(y_{k+1}-y_k)$, where $s_n$ is the partial sums of $x_n$. Then use the comparison test to argue that $$\sum _{k=1}^{\mathrm{\infty }}{s}_k(y_k-y_{k+1})$$ converges absolutely, and show how this leads directly to a proof of Abel’s Test.

Proof. We are asked to work with the sum of a product of sequences where we know at least one of them converges, so its natural to use Abel’s Lemma to rewrite this sum in terms of partial sums of the convergent sequence, and partial differences of the other. Indeed, the $n^{th}$ partial sum is (using $m=0$ in the genral form of Abel’s Lemma)

$$\sum _{k=1}^n{x}_k{y}_k=s_n{y}_n-s_0{y}_0-\sum _{k=0}^{n-1}{s}_k(y_{k+1}-y_k)$$

We wish to show that this converges as $n\to \mathrm{\infty }$. Since the right is a sum of different terms, it suffices to prove each converges and then apply the limit theorem for sums.

• We assumed $s_n=\sum _{k\le n}{x}_k$ converges. And, $y_n$ is monotone decreasing and positive, meaning its bounded below by zero. Thus monotone convergence implies $y_n$ converges, so the $s_n{y}_n$ converges by the limit theorem for products.
• $s_0{y}_0$ is a constant
• We show $\sum _{k=0}^{n-1}{s}_k(y_{k+1}-y_k)$ converges by proving it converges absolutely. Since $s_n=\sum _{k\le n}{x}_k$ converges, its a bounded sequence, so let $M$ be such that $|s_n|<M$ for all $n\in \mathbb{N}$. Thus $|s_k(y_{k+1}-y_k)|=|s_k||y_{k+1}-y_k|<M|y_{k+1}-y_k|$ for all $k$. Since $y_k$ is monotone decreasing, we know $y_{k+1}<y_k$ so $|y_{k+1}-y_k|=y_k-y_{k+1}$. This gives us an upper bound for the partial sum $$\sum _{k\le n}|s_k(y_{k+1}-y_k)|\le \sum _{k\le n}M(y_k-y_{k+1})$$ Pulling out the constant $M$ we see the sum telescopes, with $$\sum _{k\le n}{y}_k-y_{k+1}=(y_0-y_1)+(y_1-y_2)+\cdots +(y_n-y_{n+1})=y_0-y_{n+1}$$ We’ve already established the sequence $y_n$ converges, so $y_0-y_{n+1}$ tends to a finite limit $L$. This is exactly what we need to appyl comparison: the sum we care about is bounded above by a convergent sum, so $\sum |s_k(y_{k+1}-y_k)|$ converges.

Exercise 7 (Summing Cubes) Prove that the following formula holds for the sum of cubes $$1^3+2^3+\cdots +n^3={(1+2+\cdots +n)}^2$$

Hint: follow the suggested steps below:

• Let $a_k=k^2$ and $b_k=(k-1{)}^2$, and apply summation by parts.
• Simplify the left side with algebra, and the sum of squares
• Divide both sides by 4, and recognize the right as the square of what we got when summing $\sum _{k=1}^{n}k$.

Proof. We begin by applying summation by parts. Let $a_k=k^2$ and $b_k=(k-1{)}^2$.

We compute the differences: $$a_{k+1}-a_k=(k+1{)}^2-k^2=2k+1,$$ $$b_{k+1}-b_k=k^2-(k-1{)}^2=k^2-(k^2-2k+1)=2k-1.$$

Apply the summation by parts formula: $$\sum _{k=1}^n(a_{k+1}-a_k)b_{k+1}+\sum _{k=1}^n{a}_k(b_{k+1}-b_k)=a_{n+1}{b}_{n+1}-a_1{b}_1.$$

Substituting in, we get: $$\sum _{k=1}^n(2k+1)k^2+\sum _{k=1}^n{k}^2(2k-1)=(n+1{)}^2{n}^2-1^2\cdot 0^2=n^2(n+1{)}^2.$$

The left-hand side simplifies as follows: $$\sum _{k=1}^n[(2k+1)+(2k-1)]k^2=\sum _{k=1}^{n}4k\cdot k^2=\sum _{k=1}^{n}4k^3.$$

Thus, $$\sum _{k=1}^{n}4k^3=n^2(n+1{)}^2.$$

Now dividing both sides by 4

$$\sum _{k=1}^n{k}^3=\frac{n^2(n+1{)}^2}{4}.$$

This is the square of the formula for the sum of the first $n$ integers: $${\left(\sum _{k=1}^{n}k\right)}^2={\left(\frac{n(n+1)}{2}\right)}^2=\frac{n^2(n+1{)}^2}{4}.$$

Therefore, $$1^3+2^3+\cdots +n^3={(1+2+\cdots +n)}^2.$$