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.

5 0

4 years ago

2 Question Progress Homewo Simplify X-9 x² – 3x

love history [14]

Answer:

(x - 9) is already simplified

x² - 3x simplified is x(x - 3)

Step-by-step explanation:

We need to see if we can either take out GCF or factor. Since the 1st expression we can do neither, it is in its simplest form. For the 2nd expression, we can take out an x, and we get x(x - 3) as our simplified expression.

6 0

3 years ago

Differentiate y=xe^(-x/2).

Liono4ka [1.6K]

We will use the product rule:
d/dx( f(x) · g(x) ) = f´( x ) g( x ) + g´( x ) f( x ) ( and after that the chain rule )
y´ = ( x ) ´· e^(-x/2) + ( e^(-x/2)´ · x) =
= $e^{- \frac{x}{2} } + x * e^{- \frac{x}{2} } *(- \frac{1}{2})= \\ e^{- \frac{x}{2} } - \frac{x}{2} * e^{- \frac{x}{2} } = e ^{- \frac{x}{2} } (1- \frac{x}{2} )$

4 0

3 years ago