Is there a limit in this progression? 1.1 + 2.1 + 2 + 3.1 + 2 + 3 + 4

- osanseviero

Is there a limit in this progression? 1.1 + 2.1 + 2 + 3.1 + 2 + 3 + 4

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- schrodinger

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- osanseviero

Infinite? and how to demostrate it?

- hartnn

which progression is the sequence in ?

- 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...

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## More answers

- osanseviero

I cant find any order...

- hartnn

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

- hartnn

which is n(n+1)/2

- hartnn

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

- hartnn

how do you find whether next sequence has a limit ?

- osanseviero

Why is that the general term?

- osanseviero

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

- hartnn

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

- 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

- osanseviero

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

- 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

- osanseviero

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

- hartnn

so you want to find the sum ?

- hartnn

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

- osanseviero

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

- hartnn

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

- osanseviero

then how to get the limit?

- hartnn

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

- osanseviero

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

- hartnn

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

- 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

- osanseviero

What means converges?

- hartnn

sum converge means sum = finite

- 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?

- hartnn

yes, it is infinite

- osanseviero

Thanks :)

- osanseviero

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

- 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)

- hartnn

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

- 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.

- hartnn

sorry for the typos :P

- osanseviero

Thanks for a lot

- hartnn

welcome ^_^

- osanseviero

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

- osanseviero

that is an =

- 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...

- osanseviero

ok, let me try

- osanseviero

but I dont have a1

- 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

- hartnn

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

- osanseviero

nope :/

- osanseviero

So can it be determined with the things given?

- hartnn

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

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