still pewter
translation: determine and give reason as to why the following series is either convergent or divergent
clue: show first that the series is positive by analyzing the function f(x) = x - ln(1 + x) for x > 0. use maclaurin expansion to find an appropriate series that the given series can be compared to
devout nestBOT
translation: determine and give reason as to why the following series is either ...
+close
solemn sun
Do you remember the series for ln(1 + x) around x = 0?
still pewter
Do you remember the series for ln(1 + x) around x = 0?
ln1 = 0
solemn sun
Well, yes, but do you remember the series?
still pewter
mauclaurin expansion?
solemn sun
Yes.
still pewter
we are given this
it is allowed to look at
Pn(x) that approximates ln(1+x)
@solemn sun
solemn sun
Yes.
Now, we have 1/n -> 0 as n -> ∞. So:
1/n - ln(1 + 1/n) = 1/n - (1/n - 1/(2n^2) + o(1/n^2)) = 1/(2n^2) + o(1/n^2), n -> ∞
So, for large n 1/n - ln(1 + 1/n) behaves roughly like 1/(2n^2).
And this is easier to analyze.
still pewter
Yes.
Now, we have 1/n -> 0 as n -> ∞. So:
1/n - ln(1 + 1/n) = 1/n - (1/n - 1/(2n...
hmm
i was gonna ask about o(1/n^2) but if i recall correctly, by definition, o(1/n^2) = some function expressed through n such that the limit of this function/(1/n^2) = 0
meaning that we do not have to worry about o(1/n^2)
now we have 1/2n^2 like you said which ~ 1/n^2 which is comparable to a p-series with p > 1 --> convergence
thus, original series converges
solemn sun
Yes, very good!
solemn sun
<@290136354233384960>btw why is the expression inside of the o = 1/n^2
The other terms have higher order than 1/n^2 for n -> ∞.
solemn sun
Yes.
still pewter
@solemn sunalso, can't we just look at the original series
and say that it converges instantly
because all 1/n -> 0
and we have ln1 left
in the limit
surreal joltBOT
apex predator
hearty turret
<@290136354233384960>also, can't we just look at the original series
nah
still pewter
@hearty turret go back to pl server
hearty turret
hearty turret
wheeler macaroni constant is a little based
still pewter
macaroni lol
hearty turret
euler-mascheroni is no fun
solemn sun
because all 1/n -> 0
That is only a necessary condition for convergence, not sufficient.
undone galleon
Was this solved and closed? I found a nice solution that only requires you to know that log is the integral of 1/x
solemn sun
Was this solved and closed? I found a nice solution that only requires you to kn...
Well, if it was closed, you wouldn't be able to write here 😅
What did you have in mind?
undone galleon
Look at the partial sums of 1/n and log(1+1/n) separately. Summing the first n terms of log is log(n+1). You can estimate log(n+1) <= 1/1 +1/2 + ... + 1/n by using a Reimann sum for 1/x. The partition for this consists of intervals [k, k+1] for k = 1, ... , n+1, and the associated function value is taken at the left end point 1/k.
Likewise, you have 1/2 + 1/3 + ... + 1/(n+1) <= log(n+1) by doing almost the same thing but taking the value of 1/x at the right end point of [k,k+1].
From there, do some algebra and apply the monotone convergence theorem.
solemn sun
Hm, yeah, I suppose that would work.
Though, that's quite a complicated approach for this kind of problem 😅
undone galleon
I think it's intuitive. If you draw the picture, it's much simpler than reading it.
The proof above works, but it doesn't really tell you much. This gives you a way to understand why the two series are closely related
solemn sun
Oh, no, I see what you mean. It's just when I see f(x) and ln(1 + f(x)) for f(x) -> 0, you can provably guess what my first idea would be.
undone galleon
That's fair. I would want to avoid all the terms you get with expanding log because it looks like a nightmare. You have, for each n, a power series and then you sum up each power series. It's not obvious to me that will converge