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Here are some questions about series and functions. The task is to provide a counterexample for false statements and a proof for true statements (which are at most two).

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-(I) Let (a$_n$)$_n$$_\in$$ _\Bbb N$ and (b$_n$)$_n$$_\in$$ _\Bbb N$ be two sequences of real numbers such that $\sum_{n=1}^\infty (a_n)$ converges and $\sum_{n=1}^\infty (b_n)$ diverges to positive infinity. Then:

  1. $\sum_{n=1}^\infty sin(a_n^2)$ converges.
  2. $\sum_{n=1}^\infty \frac 1{(1+b_n^2)}$ converges.
  3. $\sum_{n=1}^\infty \sqrt[]{|a_n|}(b_n^2)$ diverges.
  4. $\sum_{n=1}^\infty (-1)^na_n$ converges.

-(II) Consider $f,g: \Bbb R\rightarrow \Bbb R$. let $f$ be continuous and have an absolute minimum. Also, let $g$ be bounded and have an absolute minimum. Then:

  1. $g\circ f$ is continuous.
  2. $f\circ g$ is bounded.
  3. $g\circ f$ has an absolute maximum.
  4. $f$ is bounded.
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For question (I), all options are incorrect, so for contradicting options (1) and (4), take aₙ = (-1)ⁿ/√n, for option (2) take bₙ= 1/n, for option (3) take aₙ=1/n² and bₙ=1/n.
For question (II), option (a) is incorrect, take,$g(x)= \begin{cases} 1, & \text{if $x$ is rational} \\ -1, & \text{if $x$ is irrational} \end{cases}$. And take f(x)=x².
Option (b) is correct, since g is bounded on $\mathbb{R}$, and since, f(x) is continuous on $\mathbb{R}$ which is restricted on bounded domain g($\mathbb{R}$), so fg must bounded on $\mathbb{R}$.
option (c) is correct, since g is bounded on whole $\mathbb{R}$. So restricted g on f($\mathbb{R}$) must be bounded. So, g can take absolute maximum value.
Option (d) is incorrect, take f(x)=x²

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  • $\begingroup$ Ok, thanks a lot. Can you clarify me how can you solve options 2, 3 and 4 (I) please. $\endgroup$ – TorriCinesi Jul 6 '20 at 15:48
  • $\begingroup$ For option (2), if I put bₙ=1/n, then the series become ∑n²/(1+n²), clearly as n tends to ∞, n²/(1+n²) tends to 1, so clearly this is a divergent series( since necessary condition for a convergent series ∑aₙ is lim aₙ=0). For option (3), if I put bₙ=1/n and aₙ=1/n², then series becomes ∑1/n³ which is a convergent series. For option (4) if I put aₙ=(-1)ⁿ/√n, then series becomes ∑1/√n , which is a divergent series. $\endgroup$ – A learner Jul 6 '20 at 16:15

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