MHB Does the Series Converge at \( \sqrt{2} \) and \( -\sqrt{2} \)?

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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?
 
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tmt said:
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.
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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