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

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

Mathematics
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Infinite? and how to demostrate it?
which progression is the sequence in ?
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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I cant find any order...
each term is just the sum of '1st n natural numbers'
which is n(n+1)/2
thats your general term \(a_n =n(n+1)/2\)
how do you find whether next sequence has a limit ?
Why is that the general term?
a1=1(1+1)/2=2/2=1 Does not apply...
1st term is indeed 1 right ? 1, 1+2 , 1+2+3 , .... 1,3,6, 10,...
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
Oh...now I understand! Give me a minute
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
So: 2Sn= (n+1) + (n+2) + (n+4)...ehm...this is not going right
so you want to find the sum ?
\(\Large (1/2)\sum (n^2+n) = (1/2)\sum n^2+(1/2)\sum n=...?\)
Oh...I thought that I could add Sn + Sn, and then divide them by two to get the general term
possible but i don't think its that easy....
then how to get the limit?
how do you find whether next sequence has a limit ? i may know it, but not in these "limit of next sequence" terms ...
When it says "the next sequence" it refers to the one I put
ok, so you just want to find whether the sum converges or not ?
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
What means converges?
sum converge means sum = finite
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?
yes, it is infinite
Thanks :)
How did you wrote the general term? Is there a mathematical way to get it?
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)
so, sum of 1st n natural numbers = n (n+1)/2
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.
sorry for the typos :P
Thanks for a lot
welcome ^_^
\[an?\left\{ an=\frac{ 1 }{ 2 }\left( a _{n-1} + \frac{ 10 }{ a _{n-1} }\right) \right\} for n \ge2\]
that is an =
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...
ok, let me try
but I dont have a1
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
yes, i though it was already given....
nope :/
So can it be determined with the things given?
i am not sure about this....can you ask in new post so that others can try.....sorry.

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