Finding if a series converges

[tmt1]

I have this series

$$\sum_{n =0}^{\infty}\frac{(-1)^n {x}^{2n}}{{2}^{n + 1}}$$

I need to find whether it converges or diverges at $\sqrt{2}$ and $-\sqrt{2}$.

I'm not quite sure how to approach this. For $\sqrt{2}$ I have

$$\sum_{n =0}^{\infty}\frac{(-1)^n {\sqrt{2}}^{2n}}{{2}^{n + 1}}$$Which I suppose would equal

$$\sum_{n =0}^{\infty}\frac{(-1)^n {2}^{n}}{{2}^{n + 1}}$$

Then

$$\sum_{n =0}^{\infty}\frac{(-1)^n }{2}$$

I suppose this diverges since the limit of $\frac{1}{2}$ as $n$ approaches zero is not zero, therefore it diverges.

For $- \sqrt{2}$, I suppose I have

$$\sum_{n =0}^{\infty}\frac{(-1)^n {(- \sqrt{2})}^{2n}}{{2}^{n + 1}}$$

I'm not sure how to evaluate ${(- \sqrt{2})}^{2n}$. Would it be $(-1)^{2n} \cdot 2^n$?

So I have :

$$\sum_{n =0}^{\infty}\frac{(-1)^n (-1)^{2n} {(\sqrt{2})}^{2n}}{{2}^{n + 1}}$$

Then

$$\sum_{n =0}^{\infty}\frac{(-1)^n (-1)^{2n} {2}^{n}}{{2}^{n + 1}}$$

And

$$\sum_{n =0}^{\infty} (-1)^n (-1)^{2n} \cdot \frac{1}{2} $$

$(-1)^{2n}$ should always be positive since $2n$ will always be even.

Therefore, we have

$$\sum_{n =0}^{\infty} (-1)^n \cdot \frac{1}{2} $$

Which should be divergent since the limit of $\frac{1}{2}$ as n approaches infinity does not equal 0 and $(-1)^{n}$ is alternating.

Is this correct?

 

[Prove It]

I have this series

$$\sum_{n =0}^{\infty}\frac{(-1)^n {x}^{2n}}{{2}^{n + 1}}$$

I need to find whether it converges or diverges at $\sqrt{2}$ and $-\sqrt{2}$.

I'm not quite sure how to approach this. For $\sqrt{2}$ I have

$$\sum_{n =0}^{\infty}\frac{(-1)^n {\sqrt{2}}^{2n}}{{2}^{n + 1}}$$Which I suppose would equal

$$\sum_{n =0}^{\infty}\frac{(-1)^n {2}^{n}}{{2}^{n + 1}}$$

Then

$$\sum_{n =0}^{\infty}\frac{(-1)^n }{2}$$

I suppose this diverges since the limit of $\frac{1}{2}$ as $n$ approaches zero is not zero, therefore it diverges.

Correct, also the series itself oscillates between 0 and 1/2, which is another reason why it is divergent.

For $- \sqrt{2}$, I suppose I have

$$\sum_{n =0}^{\infty}\frac{(-1)^n {(- \sqrt{2})}^{2n}}{{2}^{n + 1}}$$

I'm not sure how to evaluate ${(- \sqrt{2})}^{2n}$. Would it be $(-1)^{2n} \cdot 2^n$?

So I have :

$$\sum_{n =0}^{\infty}\frac{(-1)^n (-1)^{2n} {(\sqrt{2})}^{2n}}{{2}^{n + 1}}$$

Then

$$\sum_{n =0}^{\infty}\frac{(-1)^n (-1)^{2n} {2}^{n}}{{2}^{n + 1}}$$

And

$$\sum_{n =0}^{\infty} (-1)^n (-1)^{2n} \cdot \frac{1}{2} $$

$(-1)^{2n}$ should always be positive since $2n$ will always be even.

Therefore, we have

$$\sum_{n =0}^{\infty} (-1)^n \cdot \frac{1}{2} $$

Which should be divergent since the limit of $\frac{1}{2}$ as n approaches infinity does not equal 0 and $(-1)^{n}$ is alternating.

Is this correct?

Correct from the divergence test as the terms don't go to 0, and again because the series oscillates between 0 and 1/2.