Answer :
Answer:
This series diverges.
Step-by-step explanation:
In order for the series to converge, i.e. [tex]\lim_{n \to \infty} a_n =A[/tex] it must hold that for any small [tex]\epsilon[/tex]>0, there must exist [tex]n_0\in \mathbb{N}[/tex] so that starting from that term of the series all of the following terms satisfy that [tex]|a_n-A|<\epsilon\ , \ n>n_0[/tex] .
It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with [tex]a_n=1[/tex] , the other is constant with [tex]a_n=3[/tex] , and the third one has terms that are approaching infinity.
Really, we can write this series like this:
[tex]a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}[/tex]
If we denote the first series as [tex]b_n=1[/tex], we will have that [tex]\lim_{k \to \infty} b_k=1[/tex].
The second series is denoted as [tex]c_k=2^k[/tex] and we have that [tex]\lim_{k \to \infty} c_k=+\infty[/tex].
The third sub-series [tex]d_k=3[/tex] is a constant series and it holds that [tex]\lim_{k \to \infty} d_k=3[/tex].
Since those limits of sub-series are different, we can never find such [tex]n_0\\[/tex] so that every next term of the entire series is close to one number.
To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose [tex]\epsilon =1[/tex], all those terms associated with the third sub-series will be out of this interval [tex](A-1, A+1)=(0, 2)[/tex].
Therefore, the observed series diverges.