## osanseviero Group Title Is there a limit in this progression? 1.1 + 2.1 + 2 + 3.1 + 2 + 3 + 4 11 months ago 11 months ago

1. osanseviero

Infinite? and how to demostrate it?

2. hartnn

which progression is the sequence in ?

3. osanseviero

The question: 1. Determine if the next sequence has a limit (demostrate it) and determine it (hint:find the general term) c) 1,1 + 2,1 + 2 + 3,1 + 2 + 3, + 4...

4. osanseviero

I cant find any order...

5. hartnn

each term is just the sum of '1st n natural numbers'

6. hartnn

which is n(n+1)/2

7. hartnn

thats your general term $$a_n =n(n+1)/2$$

8. hartnn

how do you find whether next sequence has a limit ?

9. osanseviero

Why is that the general term?

10. osanseviero

a1=1(1+1)/2=2/2=1 Does not apply...

11. hartnn

1st term is indeed 1 right ? 1, 1+2 , 1+2+3 , .... 1,3,6, 10,...

12. hartnn

and each term is sum of 1st n natural numbers 1st term = 1 2nd term = sum of 1st 2 natural numbers =1+2 =3 and so on

13. osanseviero

Oh...now I understand! Give me a minute

14. osanseviero

Cant you add the series plus iteslf? like this: Sn= 1 + 3 + 6 + 10 + ... + n-2 + n-1 + n Sn= n + n-1 + n-2 +...+ 6 + 3 + 1

15. osanseviero

So: 2Sn= (n+1) + (n+2) + (n+4)...ehm...this is not going right

16. hartnn

so you want to find the sum ?

17. hartnn

$$\Large (1/2)\sum (n^2+n) = (1/2)\sum n^2+(1/2)\sum n=...?$$

18. osanseviero

Oh...I thought that I could add Sn + Sn, and then divide them by two to get the general term

19. hartnn

possible but i don't think its that easy....

20. osanseviero

then how to get the limit?

21. hartnn

how do you find whether next sequence has a limit ? i may know it, but not in these "limit of next sequence" terms ...

22. osanseviero

When it says "the next sequence" it refers to the one I put

23. hartnn

ok, so you just want to find whether the sum converges or not ?

24. osanseviero

Look, I have 5 exercises 1. $\left( -1 \right)^{n}\left( \frac{ 5n+4 }{ 2n } \right)$ For this one we saw that the 4 will be nothing in comparison so the limit without the -1n is 5/2. It has no limit because it is ocilating. 2. $2^{-1}\times2^{2}\times2^{-3}...$ Because there is a negative sign, it will change, so there is no limit

25. osanseviero

What means converges?

26. hartnn

sum converge means sum = finite

27. osanseviero

I want to know if it has a limit and to demostrate and determine it. I think that for this infinite is the limit, am I right?

28. hartnn

yes, it is infinite

29. osanseviero

Thanks :)

30. osanseviero

How did you wrote the general term? Is there a mathematical way to get it?

31. hartnn

the sum of 1st n natural numbers ? that you can find by your method, Sn = 1+2+3+....n Sn = n+n-1 +n-2 +...3+2+1 2Sn = (1+n)+ (1+n) +..... so, each pair sum is (1+n) and there are n terms so, 2Sn = n (n+1)

32. hartnn

so, sum of 1st n natural numbers = n (n+1)/2

33. hartnn

that became your general term and for your sequence, you need to sum this, and whne n becomes large the sum goes on increaseing which menas this progression has no limit.

34. hartnn

sorry for the typos :P

35. osanseviero

Thanks for a lot

36. hartnn

welcome ^_^

37. osanseviero

$an?\left\{ an=\frac{ 1 }{ 2 }\left( a _{n-1} + \frac{ 10 }{ a _{n-1} }\right) \right\} for n \ge2$

38. osanseviero

that is an =

39. hartnn

have you solved such problems, with a (n-1) , before.... because i just could think of finding, a1,a2,a3...and see whether there's any pattern...

40. osanseviero

ok, let me try

41. osanseviero

but I dont have a1

42. osanseviero

this is just telling you that the new value is the last one added by 10 divded by iteslf $\left( a ^{2} +\frac{ 10 }{ a ^{2} }\right)$ divided by two

43. hartnn

yes, i though it was already given....

44. osanseviero

nope :/

45. osanseviero

So can it be determined with the things given?

46. hartnn

i am not sure about this....can you ask in new post so that others can try.....sorry.