How to solve limits like this

>>8977041
logically, using the power rule, it's 1

>>8977041

S(∞) doesn't make sense -- in particular it assumes that the limit exists and it may not. You could go definitional:

Let

[math] s = \mathrm{lim}_{n \rightarrow \infty} \sum f(n) [/math]

and then take S(n)/s -- with the explicit assumption that the limit exists.

Alternatively you can replace

S(n)/S(∞)

with a more concrete form like

S(n)/S(n^2)

or

S(n)/S(n!)

or even use a limit over n and m, and have

S(n)/S(n+m)

>>8977049
I would have written s(∞) in the correct limit form but im lazy. I am assuming that the limit does exist and im interested in why some functions for f(m) seem to make it approach some number or 0.
if I use 1/m! for f(m) it seems to go to 0 but if I use 1/m^2 it approaches some number near 0.608

[eqn] \begin{matrix}
\mathtt{L'} & \mathtt{H} & \mathtt{\hat{O}} & \mathtt{P} & \mathtt{I} & \mathtt{T} & \mathtt{A} & \mathtt{L} \\
\mathtt{H} & & & & & & & \\
\mathtt{\hat{O}} & & & & & & & \\
\mathtt{P} & & & & & & & \\
\mathtt{I} & & & & & & & \\
\mathtt{T} & & & & & & & \\
\mathtt{A} & & & & & & & \\
\mathtt{L} & & & & & & & \\
\end{matrix}
[/eqn]

>>8977090
how do i use l'hopital's rule when the upper bound of the summation is the variable

>>8977109
>use induction to find a closed form for s(n)
>find upper/lower bounds and use squeeze theorem
>use mean-value theorem to rewrite as s(n)=n*f(m)
chose your favorite method
there's no general algorithm that works for every kind of f, otherwise all of probability would become trivial.

>>8977041
You need some more assumptions because even if s(inf) exists and is nonzero the limit may fail to exist overall. Let f= 1/n^2. Then in the brackets we will have n*(1- 6/(pi*n^2)), which diverges to infinity.

>>8977126
Ok well I fucked that shit up it should have been a sum of 1 + 1/4 + 1/9... up to 1/n^2 times 6/pi but it will diverge anyways.
n - 6/pi*(n + n/4 + n/9 ... +1/n), goes to negative infinity clearly

>>8977133
It should be 6/pi^2 not 6/pi. Hoever it should still diverge to negative infinity

There is a a L'Hopital's rule for sequences. lim sn/tn = lim ( sn-s_(n-1) ) / (tn - t_(n-1)).

If you put 1/n in the denominator instead of n in the numerator, you get 0/0, so you can take L'Hopital and get this:

(n-1)nf(n) / S(infinity), so you only have to compute the limit of (n-1)nf(n).

>>8977265
Actually nvm, applying this theorem (it is called Stolz–Cesàro theorem) needs some things to hold first. I dunno if they do, cause I don't even know the full statement of the theorem.

>>8977304
Actually, I think it can be applied
https://math.stackexchange.com/questions/599204/stolz-cesaro-theorem-0-0-case

>>8977315
Not without other assumptions. In this case we have lim n-> inf of (S(n)-S(n+1))/(S(inf)*(1/(n+1) - 1/n)) = f(n+1)*(n+1)*n/S(inf). Its easy to see the hypothesis would fail if f was something like n^(-3/2)

>>8977133
Not so clearly. In the case of 1/n^2 it actually works as seen by this: >>8977315

>>8977335
Well yeah I assumed that the limit (n-1)nf(n) exists. Can't think of how to approach it if all we know is that Σf(i) converges.

So does anybody know what happens in the general case where we don't know if the limit of (n-1)nf(n) exists?

>>8977914
Theres no result that holds for just assuming s(inf) exists and is nonzero. f(n)=n^(-3/2) fails to have the sequence converge.

>>8979298
oh right
I'm an idiot