A Viewer-Suggested Problem on Sum of Logarithms

[@willbishop1355]

Correct answer, but an error in logic at 3:00. Both the sum from 1 to infinity of log(n) and the sum from 3 to infinity of log(n) are infinite, and you can't subtract infinity from infinity. You have to subtract term by term inside the summation. Similar problem at 5:25. In general the explanation is a bit confused in this video.

[@spiderjerusalem4009]

converting it to product would've been way easier

[@harshvadher9234]

Guys actually there is co incidence here You see this summation expansion gets all terms inside log is multiplied and this very multiplied thing came in Putnam math 1955 it's sol.=2/3 this taking log is log(2/3) 👍

[@sea34101]

You can't split divergent sums. You need to look at the partial sum, cancel the intermediate terms then take the limit. Your reasoning is incorrect.

[@hamza201183]

I think so too.

[@mcwulf25]

Even if the original summation is convergent?

[@sea34101]

@@mcwulf25 It's as if you were writing x=x*0/0

[@hamza201183]

@@mcwulf25 Yes!

[@PhysicsProf1969]

It's interesting that this method (which I agree is incorrect) gives the correct answer to the problem! That happens a lot when you play with divergent series, it seems.

[@mars_titan]

Exactly! I too observed it

[@piraterblx3119]

yep!

[@hamza201183]

Hello, are you allowed to separate the sum into multiple non-convergent sums?

[@richardfredlund3802]

presumably that's why he mentions that the series can be shown to converge by the integral test, at the beginning.

[@hamza201183]

@@richardfredlund3802 I'm not talking about the initial serie. But about the splitting into sum of clearly not-convergent series.

[@po-haochou2530]

@@richardfredlund3802 Even the original series converge to A, when you spit it into several not-convergent series , their summation may not A

[@richardfredlund3802]

@@po-haochou2530 @hamza ohk i don't know then if there is a problem there

[@hamza201183]

@@richardfredlund3802 Sure bro, no problem, thanks for the reply :) This is what maths are good for: Discussions :)

[@simonmultiverse6349]

I would immediately make the substitution: sum-of-log = log-of-product. I imagine that there would be massive cancellation in the product, so the answer would be something like log(function of 1/infinity) +- log(function of first term)

[@richardfredlund3802]

very nice...something quite satisfying about telescoping series

[@matanah1989]

in the first term the partial sum S1_N is log1+log2 - logN-log(N-1) you can't ignore that the second term the partial sum S2_N is -log3+log(N^2+N+1) together S_N = S1_N + S2_N = log1+log2-log3+log(N^2+N+1)-logN-log(N-1) = log(2/3) + log((N^2+N+1)/(N(N-1))) the argument approaches 1 so log approaches 0. so SN --> log(2/3)

[@justsomeboyprobablydressed9579]

I love all your videos. Great fun and preparation for math contests. Do you mind telling me what writing/whiteboard software you are using?

[@yoav613]

Nice and very easy

[@mcwulf25]

This is how I did it. Nice problem 👍

[@bhoomib8578]

why don't you try geometry and other topics as well we will appreciating those as well!

[@VSN1001]

I used a telescopic product instead by rewriting the expression from stigma notation to pi notation

[@piraterblx3119]

yea! its that simple. just by putting the log infront of the sigma which will convert the "stigma LOL" into the pi nottation. thats how I did it :3

[@lunstee]

Right at the opening, you say that this sum converges and explicitly omit showing so. Without supporting this, your analysis is really taking the difference of two nonconverging sums, basically infinity-infinity, which is a problem that I don't think should be dismissed quite so casually. However, it's easy enough to prove the final result without this problem by looking at the sum of a finite number of terms, then taking the limit as that sum goes to infinity. The sum from 2 to N of log((n-1)/(n+1)) is log2 - logN - log(N+1). This does not converge, so you can't really collapse this telescoping sum: as n goes to infinity, the goes is log2 - log(infinity)-log(infinity). Similarly, the sum of log((n^2+n+1)/(n^2-n+1)) is just log((N^2+N+1)/3), which similarly goes to log(infinity^2) as N goes to infinity. What lets things work is when you sum the two series (the sum of the original expression) to the same N. The finite sum is log(2/3)+log((N^2+N+1)/N/(N+1)). This can be written as log(2/3) + log((1+1/N+1/N^2)/(1+1/N)). As N goes to infinity, the 1/N and 1/N^2 terms go to zero leaving the last term as log(1/1) which is zero.

[@mohanabavi]

How the second sum is negative while the summands are positive?

[@reeeeeplease1178]

Because he isnt allowed to calculate the series like he does in the video (splitting into divergent series)

[@nirmankhan2134]

This problem was copied from that Putnam question.

[@sadnankhan3042]

Log (infinity)is not =0

[@lunstee]

but -log(infinity) -log(infinity) + log(infinity^2) does happen to be 0 😁

[@spitsmuis4772]

This is just taking the log of your earlier video ( kzbin.info/www/bejne/opDEZ4CBjtlksLc )

[@mathcanbeeasy]

This is totally wrong. From your method you obtain the series 1-1+1-1+1-1+.... as equal to 0, which is not correct. There are two partial sums, one is 1 and other is 0. An infinite series is convergent if and only if the sequence of the partial sums is a convergent sequence and the sum of the infinite series is the limit of the sequence of the partial sums. But, for the series of the first log-log, after the reducing terms like you did (I don't know why you stoped and not continued to the n) the Sn is log(1)+log(2)-log(n)-log(n+1), which is divergent. And the other sequence of partial sums for the second difference of logs. So, your solution is incorrect. I hope you return with the correct one. 🙂