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

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.