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I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I'd like to ask, if my proof to the below question on convergence of infinite series is rigorous and sufficient, and checks out. $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Give an example of each or explain why the request is impossible referencing the proper theorem(s).

(a) Two series $\sum {x_n}$ and $\sum{y_n}$ that both diverge but where $\sum x_n y_n$ converges.

(b) A convergent series $\sum x_n$ and a bounded sequence $(y_n)$ such that $\sum x_n y_n$ diverges.

(c) Two sequences $(x_n)$ and $(y_n)$ where $\sum x_n$ and $\sum (x_n + y_n)$ both converge but $\sum y_n$ diverges.

(d) A sequence $(x_n)$ satisfying $0 \le x_n \le 1/n$ where $\sum (-1)^n x_n$ diverges.

Proof.

(a) The simplest examples I could come up with are:

(i) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent.

(ii) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n+1}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.

To see that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent, we simply note that $\sum_{n=1}^{\infty}\frac{1}{n^p}$ is convergent for $p > 1$, so $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent. And $\frac{1}{n(n+1)} < \frac{1}{n^2}$, so by the Comparison test, $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.

(b) I think that this request is impossible. The series $\sum x_n$ is convergent. By the Cauchy Criterion, given any $\epsilon > 0$, there exists $N \in\mathbf{N}$, such that \begin{align*} \absval{x_{m+1} + x_{m+2} + \ldots + x_{n}} < \frac{\epsilon}{M} \end{align*} for $n > m \ge N$.

Also, the sequence $(y_n)$ is bounded, so $\absval{y_n} \le M$ for all $n\in\mathbf{N}$.

Consider the expression $\absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}}$. We can write, \begin{align*} \absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}} &\le \absval{x_{m+1}\absval{y_{m+1}} + \ldots + x_{n}\absval{y_{n}}}\\ &\le \absval{x_{m+1}M + \ldots + x_{n}M}\\ &=M \absval{x_{m+1} + \ldots + x_{n}}\\ &<M \cdot \frac{\epsilon}{M} = \epsilon \end{align*}

So, by the Cauchy criterion, $\sum x_n y_n$ is a convergent series.

(c) This request is impossible. By the Algebraic Limit Theorem, if $\sum (x_n + y_n)$ converges and $\sum y_n$ converges, then $\sum (x_n + y_n) - \sum (y_n) = \sum x_n$ is also convergent.

(d) I think that this request is impossible as well. We know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges. I have a hunch, that the the alternating series $\sum_{n=1}^{\infty}(-1)^n x_n$ is always bound by $\sum_{n=1}^{\infty}(-1)^n /n$.

By the Algebraic Limit theorem, $\lim_{n \to \infty} 0 \le \lim_{n \to \infty} \le \lim_{n \to \infty} \frac{1}{n}$, so $\lim_{n \to \infty}x_n = 0$. I would like to show that $(x_n)$ is a decreasing sequence.

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    $\begingroup$ For b), look at alternating series. $\endgroup$ – David Mitra Dec 30 '20 at 14:27
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    $\begingroup$ In $(d)$, consider a sequence that is $x_{2n}=\frac 1{2n},x_{2n+1}=0$. $\endgroup$ – abiessu Dec 30 '20 at 14:30
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    $\begingroup$ i think (c) is reasoned incorrectly: it's not that $\sum (x_n + y_n) - \sum (y_n) = \sum x_n$ for whatever x and y and then $\sum$ x is also convergent. it's that because $\sum$ y and $\sum$ x+y converge that we say that $\sum (x_n + y_n) - \sum (y_n)$ is equal the convergent series $\sum x_n$... or something like that. i think i'll just do some contrapositive though: $\sum (x_n + y_n)$ diverges if $\sum (x_n)$ or $\sum (y_n)$ diverges. let's see what José Carlos Santos has to say $\endgroup$ – BCLC Dec 30 '20 at 15:34
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    $\begingroup$ @BCLC, for (c) I meant to use the result : if $\sum a_n = A$ and $\sum b_n = B$, then $\sum (c_1 a_n \pm c_2 b_n) = c_1 A + c_2 B$. $\endgroup$ – Quasar Dec 30 '20 at 15:41
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(a) What you did is fine.

(b) $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n$ converges, but $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n\times(-1)^n$ diverges.

(c) What you did is fine.

(d) Take$$x_n=\begin{cases}\frac1n&\text{ if $n$ is odd}\\2^{-n}&\text{ otherwise.}\end{cases}$$

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    $\begingroup$ Both the first and the second inequalities are false. $\endgroup$ – José Carlos Santos Dec 30 '20 at 14:51
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    $\begingroup$ @BCLC It seems to me that what the OP did is fine: if both series$$\sum_{n=0}^\infty(x_n+y_n)\text{ and }\sum_{n=0}^\infty x_n$$are convergent, then so is the series$$\sum_{n=0}^\infty\bigl((x_n+y_n)-x_n\bigr),$$which is equal to $\sum_{n=0}^\infty y_n$. $\endgroup$ – José Carlos Santos Dec 30 '20 at 15:39
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    $\begingroup$ @JoséCarlosSantos, (d) how do I see that the series $\sum (-1)^n x_n$ with terms $x_{2n+1} = \frac{1}{n}$ and $x_{2n}=2^{-n}$ is divergent? $\endgroup$ – Quasar Dec 30 '20 at 15:50
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    $\begingroup$ The series with positive terms diverges and the series with negative ones diverges. $\endgroup$ – José Carlos Santos Dec 30 '20 at 17:24
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    $\begingroup$ @JoséCarlosSantos, I see that. And the idea is that if $\sum x_n$ is convergent, and $\sum y_n$ is divergent, $\sum (x_n + y_n)$ is divergent. But, the two series are interlaced, and we need to rearrange terms. Can I freely do that? $\endgroup$ – Quasar Dec 30 '20 at 20:27

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