Question

Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Answer 1

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$>0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that  $|a_n-A|n_0$ .

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 $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we  denote the first series as $b_n=1$, we will have that $\lim_{k \to \infty} b_k=1$.

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$.

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$.

Since those limits of sub-series are different, we can never find such $n_0$ 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 $\epsilon =1$, all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$.

Therefore, the observed series diverges.