Hi, so wolfram alpha says this series converges but in a video the organic chem tute did a series that just had a square root and he concluded that diverged. Would this cube root series be similar to a square one?
#Series diverge or converge
fair cosmos
fossil merlinBOT
Hi, so wolfram alpha says this series converges but in a video the organic chem ...
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edgy gulch
Hi, so wolfram alpha says this series converges but in a video the organic chem ...
As far as I know, this one does converge. Could you post the video by The Organic Chemistry Tutor containing the series you mentioned (the one with a square root)?
fair cosmos
https://www.youtube.com/watch?v=0YeON4p0ogw at timestamp 30min
edgy gulch
https://www.youtube.com/watch?v=0YeON4p0ogw at timestamp 30min
That one is the same as $$\sum\limits_{n=1}^\infty n^{-\frac{1}{2}}$$
void zincBOT
Rafain | #WhoIsWolf
edgy gulch
In general, the p-series $\sum\limits_{n=1}^\infty n^p$ converges if and only if $p<-1$
void zincBOT
Rafain | #WhoIsWolf
fair cosmos
oh ok does the p series not apply here?
edgy gulch
oh ok does the p series not apply here?
For the problem you posted? It does, the given series is comparable to the p-series with p = -4/3
fair cosmos
won't the series in the picture be greater than n^(-4/3)?
edgy gulch
won't the series in the picture be greater than n^(-4/3)?
That is true
How about we compare the 2nd term of the given series to the 1st term of the p-series with p = -4/3, 3rd to 2nd, etc.?
The absolute convergence of a series (or lack thereof) is not concerned with a mere, finite difference
fair cosmos
ah ok so we modify the tail of the given series to start below the start of n^(-4/3)?
edgy gulch
Or we modify the series $\left{\sum_{n=1}^k a_n: k\in\mathbb N\right}$ to be compared, to have the following sequential terms
$$a_1 = 3^{\frac{4}{3}}, a_{n+1} = n^{-\frac{4}{3}}\text{ for } n\in\mathbb N$$
fair cosmos
what does that tell us?
edgy gulch
what does that tell us?
This series converges, since its 2-tail is the p-series with p = -4/3
Every of its sequential term is no less than the corresponding term in the given series, so it can be used for the Direct Comparison Test
fair cosmos
so are we comparing the terms n=2 in both series? because isn't the given series still greater
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Rafain | #WhoIsWolf
edgy gulch
so are we comparing the terms n=2 in both series? because isn't the given series...
Are you sure? Could you write out the second sequential terms of both series?
fair cosmos
edgy gulch
$$a_2 = 1^{-\frac{4}{3}} = 1$$
void zincBOT
Rafain | #WhoIsWolf
fair cosmos
Oh because we are doing the a(n+1) term for a given n?
edgy gulch
I defined it as such, yes
Then $(a_n)^3 = (n-1)^{-4}\geq \frac{1}{n^4 - \frac{2}{3}}$ for any natural $n\geq 2$
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Rafain | #WhoIsWolf
fair cosmos
right ok I think I got it. thanks @edgy gulch
nimble flintBOT
@fair cosmos has given 1 rep to @edgy gulch
fair cosmos
+close