# Doubts about series convergence/divergence and properties of compound functions.

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).

-> Questions in image format <-

/Question in text format/

-(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.

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

• Ok, thanks a lot. Can you clarify me how can you solve options 2, 3 and 4 (I) please. – TorriCinesi Jul 6 '20 at 15:48
• 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. – A learner Jul 6 '20 at 16:15