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Let $a_n$ be a sequence of real numbers. Which of the following is true?

a. If $\sum a_n$ converges,then so does $\sum a_n ^4.$

b.If $\sum |a_n|$ converges,then so does $\sum a_n ^2.$

c.If $\sum a_n$ diverges,then so does $\sum a_n ^3.$

d.If $\sum| a_n|$ diverges,then so does $\sum a_n ^2.$

My effort: I think option $a,c$ are incorrect. Because for $a.$ if we take $a_n=(-1)^n/n^{1/4}$. (By Alternating series convergence criterion). I dont know about option $b.$ and $c.$ For option $d.$ I am doing as $|a_n|\le a_n^2$ so by applying comparison test it can be concluded $d.$ is correct. Though I am not sure.

Problem $2$: I need a little clarification about the problem:Whether the series $\sum_{n=1}^{\infty}cos(3^nx)/2^n$ converges for all $x\in \mathbb R.$

My Effort: As we take $M_n=1/2^n$ , then $\sum M_n$ converges. So I shall conclude that the series is convergent by Weirstrass M test for all $x \in \mathbb R$. Is this correct?

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user151456
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1 Answers1

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a) : Let $a_n = \frac{(-1)^n}{n^{\frac{1}{8}}}$. $\sum a_n$ converges from Abel's Test, if for nothing else. But, $b_n = a_n^4 = \frac{1}{\sqrt{n}}$ is not the general term of a converging series, since $\sum \frac{1}{n^a}$ converges iif a>1.

b) : $\sum |a_n|$ converges $\implies (|a_n|) \rightarrow 0 \implies |a_n| < 1$ for $n \geq N_o$. So : $ n \geq N_o \implies a_n^2 < |a_n| \implies \sum a_n^2 $ converges by comparison test with series of positive terms.

c) : $\sum \frac{1}{n}$ diverges, but $\sum \frac{1}{n^3}$ converges, so false.

d): Idem, $\sum \frac{1}{n}$ diverges, but $\sum \frac{1}{n^2}$ converges, so false.

In conclusion, only b) is true.

Problem 2 : let $c_n(x) = cos(3^nx)/2^n $

$|c_n(x)| \leq \ 1/2^n = d_n$

From this you get that $\sum c_n(x) $ converges absolutely by comparison of series of positive terms (this implies pointwise convergence which is, I think, what you were looking for), and uniformly (so relatively to x, as a series of functions. This is much stronger than simple pointwise convergence) because:

$|R_n(x)| = |\sum_{k=n}^{+\infty} c_n(x) | \leq \sum_{k=n}^{+\infty} |c_n(x)| \leq \sum_{k=n}^{+\infty} 1/2^n =\sum_{k=n}^{+\infty} d_n = 1/2^{n-1}$

Thus giving: $Sup(R_n(x), x\in\mathbb R) \leq 1/2^{n-1}$

That proves the uniform convergence.

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