Series converge?

1. Jun 2, 2007

chesshaha

1. The problem statement, all variables and given/known data

series converge?

DL the img, sry, can't type it.
http://www.geocities.com/chessobeyer/math.bmp

what's the sum if converge?

2. Relevant equations

none.

3. The attempt at a solution

alternating series, but it's (-2)

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Last edited: Jun 2, 2007
2. Jun 2, 2007

quasar987

Are you saying it converges, or diverge?

3. Jun 2, 2007

chesshaha

I don't know where to start, I didn't know what test to use?

Can u give me a hint of which test to use?

4. Jun 2, 2007

quasar987

Based on what you said, I assume that you already understood that the series can be rewritten as

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

??

I don't see which test can be used on this. But you can fall back on the very definition of convergence: A series converge if the sequence of the partial sums converge. But if you find two subsequences that converge to different values, then the sequence itself diverges. Can you find those subsequences?

5. Jun 2, 2007

chesshaha

thx, that helps alot, so the alternating series?

Can u factor out the 2? r u sure that can be done?i don't understand

6. Jun 2, 2007

chesshaha

I got the series diverge, because the 2 cancels and it left with $$\sum_{n=1}^{\infty}(-1)^{n}$$

is it right?

thx for ur help

7. Jun 2, 2007

quasar987

Ah, there is a test you can use! And it's the easiest. If $\lim_{n\rightarrow \infty} a_n\neq 0$, then the series $\sum a_n$ diverges.

8. Jun 2, 2007

chesshaha

Thanks for your help. But I still don't understand how can did u rewrite the series... to this $$\sum_{n=1}^{\infty}2(-1)^{n}$$

Sorry, I am a bit of slow, Please Explain more, Thank You!

9. Jun 2, 2007

quasar987

Ok, here is how.

$$\frac{(-2)^{n+1}}{2^n}=\frac{(-2)(-2)^n}{2^n}=(-2)\left(\frac{(-2)}{2}\right)^n=(-2)(-1)^n=2(-1)^{n+1}$$

Thus,

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

10. Jun 3, 2007

ice109

catching the bump up in the index is tricky. notice that the series now starts at n=1.

11. Jun 3, 2007

chesshaha

Thank you very much, this helps alot.

So the series converge, the sum is either 0 or -2, depends if it's even or odd, right?

12. Jun 3, 2007

Dick

A series cannot converge to two limits. That sort of behavior is called 'divergent'.

13. Jun 3, 2007

chesshaha

o yea, thanks.

14. Jun 4, 2007

Gib Z

Ahh I think a more appropriate word would have been is oscillating =]

15. Jun 4, 2007

Dick

Right. And the sum of i from i=0 to infinity could more accurately be said to 'increase without bound' rather than diverge.

16. Jun 4, 2007

Gib Z

Taking out the factor of 2, partial sums gives : -1, 0, -1,0,-1.... So it is bounded...