This Sum Amazes Me Every Time 

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I understand this prove for the first time. Now, I'll just have to practice a while.

It is also the special case of the Dirichlet Eta Function η(1) and also the x = 1 case of the Mercator Series

That's a very neat way of proving that the sum equals ln 2

Did this video explanation go too fast?

"Oh, good. A short video." 30 seconds later... "Ahhhhh this is going way too fast!"

Well i just remembered the Taylor expansion of ln(1 + x) = infinite sum( ((-1)^(n+1))x^n/n) (we use it a lot in calculating limits) which kinda solves the problem in a blink but I really liked that you squeezed in the squeeze theorem ;)

Okay. That limit of inequality...it blew my mind!

This is so clean! Perfect speed imo :)

I really prefer this more advanced and faster going through content

4:17 did you say "I'll sum you in that one" instead of the usual "I'll see you in that one" 😆. Love your videos btw!

for a second I was like wait, this isn't what's promised in the thumbnail, then I remembered what natural log means.

This is mind blowing !

Inequalities are a pain but sometimes they deliver the goods.

The integral of x^n over 1+x can easily be computed by reindexing, just let y=1+x, it becomes integral of (y-1)/y with altered bounds

You amaze me every time too

A cool little thing I found is that if you write the sum as 1 to inf as ((-1)^(n+1))/(xn), for every positive integer above 1 you put for x, the sum equals to the xth root of 2 (2^(1/x)). neat.

Another great video. Huge thanks. When I paused the video I realized that you were not really going too fast. I would have done the same.

Quality videos. Subbed.

Math major 15 years ago and my head hurts with all these integrals. But nice deduction!

that is actually amazing

The length of the video is a nice number

Why didn’t you just plug -t into the geometric series, integrate from 0 to x-1, and plug in 2?

Amazed..!

Brian has abandoned the dark side

Got tired of writing out your math backwards huh 😂

Your proof was quite beautiful, but simple it was the Dirichlet ETA function right??

Who also thought of Candace from Phineas and Ferb when he said "conditionally"

Awesome

Amazing!!!! OH OMG, this is amazing! I figured it for 3 days and this is my answer?! So funny👍

Epic series

Which books should I refer to learn all these ?

I like these videos a lot, but find they go a beat and a half to fast for me to track, I often need to pause to not feel totally disoriented. It's awesome content and we'll explained, just too fast for me to stay with you as you are explaining.

1:34 "We can pull this sum out of the integral with basically no justification since it's a finite sum." 3:40 takes the limit as n approaches infinity, where n was the bound of the earlier "finite" sum. ... I feel like justification might have been needed.

1:03 I think you mean divide by 1-x.

so impressed that it is 5 minutes & great video!!!!!!!!!!!!!!!!!!!!!!!!!!

Easier to take ∫ dx/(1+x) = ln(x+1). Integrate the geometric series on the left to get x-x^2/2+x^3/3 ... and substitute x=1.

Hello, a request, can you please please make a video on that integral inequality, I still don't understand the proof of it.... please

I'mJapanese.this video is very interesting!!

Love Math ))) It's THE ONLY HUMAN READABLE form of the REALITY ))) Ironically it's not reality itself, But precise DESCRIPTION of reality ))) No other forms of REALITY are available to us as we are A PART of it )))

4:20 video length Man of culture I see.

Clean video

I don't think it's too fast. If people are happy to dedicate some time, they can watch through it once, then watch through again and pause as necessary if they wanna take notes.

Nice vid

Whoa 😳

Why are u taking the primitive between 0 and 1 at 2:09 and why isnt it 0 and +inf

Consider F(x) = 1-x/2 + x^2/3 - x^3/4 + ... . Now, d(x F(x))/dx = 1- x+x^2 -x^3+ ... = 1/(1+x) . Integrating both sides x F(x) = Log(1+x)+c . Clearly, c=0 considering x=0 on both sides. Therefore, F(x) = Log(1+x)/x . The series in the video is x=1. This also shows that as x approaches -1, we get log divergence which is the divergence rate for harmonic series. More interestingly, one can use this to define analytic continuation for the similar series for x = some negative number not equal to 1.

1:37 those equations looks soooo HOT

Loved this derivation, but wouldn't it be enough to calculate it through taylor series of log(1+x) ?

Looking at those exponents made me realize something to the power of 1/2 means square root, which means if you take the negative of a number and take it to the power of one half it becomes an imaginary number. Just tried it on my phone and you need parentheses but it said, keep it real rather than error. Thought it was funny.

I tried solving the alternating harmonic series. I took it to be the summation of (-1)^n/(n+1). And I write (-1)^n as e^n(ln-1) and I wrote ln(-1) as iπ and then I used Euler's identity and wrote is as the summation of (cos (nπ)+isin(nπ))/n+1 and then on breaking the summation the the summation which contains sin(nπ) becomes zero. And we get summation of (cos(nπ))/n+1. But I don't know what to do next. I'm stuck on this from days. I tried solving it like this, but I'm stuck....go nowhere, thanks by the way...... amazing video.....

isnt the sum from k=1 to infinity} ((-1)^{k+1 of)/k a convergent series?

Why this Channel has only 62000???

is the reindexing at the end of the video correct?

was sagt er am Schluss?

Just use Taylor series of ln(1+x) at x=1

4:20

I mean you could probably just taken the Taylor series of ln(x)?

this is before watching, are you saying that such a simple sum can equal the nautral log of 2?????

Nice

Another approach to the problem. We know that the sum of the ∞ Geometric Series, 1-x+x²-x³... = 1/(1+x). Taking the definite integral from 0 to 1 on both sides, we get: x-(x²/2) +(x³/3)....∞ = ∫1/(1+x). Solving the RHS integral, we get: ln(1+x) = x-(x²/2) +(x³/3)....∞ The definite integral is from 0 to 1, and plugging in these values in the LHS and RHS, we get: ln2 - ln1 = [1 - (1/2) + (1/3) - (1/4).....∞] - [ 0-0+0-0....] Hence, ln2 = 1 - (1/2) + (1/3) - (1/4).....∞ Taking log on both sides, we get: e ^ (1 - (1/2) + (1/3) - (1/4).....∞) = 2 Cheers!

0:18 But that's not generally true, right?

Just the Taylor series for ln(2).

You can explain this much better, if you just integrade the geometric series expansion and plug in 1 as value.

you squeeze this in 4 mins :)))

At 3:06 isn't x to the power of n only >0? instead of >=0

Me at the end of vedio : wow

I was so amazed even i want to give you 10 likes!!🤭🤭🤭

Love from India 🇮🇳🇮🇳 🇮🇳 🇮🇳 ❤❤

not really surprising, just taylor series of ln(1+x)

keep on I also try to make animated content to help students.

taylor series pops out from anywhere brother

Another crazy stuff. How can integer fractions come together as ln(2)? I think I will quit maths

If you say so.....

Applying the triangle inequality, before dropping the absolutes is very daring of you :p

ln(2)-ln(2)=0

Why do you say calc 2 at the beginning? To me at least, calc 2 is calculus on R^n, while everything you covered here is covered in calc 1

Google Translate. I'm a Japanese viewer, but I would appreciate it if you could add English subtitles.

wow nc vid

お、おう めっちゃ複雑な計算なので目でおいきれんかった😯 結論だけわかる😅

It would be rather easy with MCLAREN SERIES as according to this Log(1+x) = x- x²/2 + x³/3 - ...... Let it be eq. 1 Now Log Sn = 1 - 1/2 +1/3 -......... Log e Substituting x = 1 from eq.1 Log Sn = log (1+1) x 1 Sn = 2 :)

2:38 This doesn’t convince me. Why does this work?

e^(alternating series) can be written as e^(positive terms) / e^(negative terms). If you pull out a half from e^(negative terms) you get 1/2 e^(positive terms), which gives e^(positive terms) / (1/2) e^(positive terms) = 1 / (1/2) = 2. QED *trollface*

gotta admit 3b1b did a simpler proof

You are cool

Who else computes alongside his videos ? .......(and gets stuck all the time 😑)

Nice video. We can also use conditionally convergent series to show why the algebraic 'proof' of 0.999... = 1 is invalid. (By the way, did you see my comment under your last video where I explained why your formal argument for 0.999... = 1 is flawed?) The algebraic proof is where we start with x = 0.999..., we multiply both sides by 10, subtract what we started with, then divide both sides by 9 and supposedly end up with x = 1. Obviously if we start with x = 0.999... and then multiply both sides by (10 - 1)/9 we should end up with exactly what we started with. The reason it appears to change to x = 1 is because under the cover of the decimal system a sneaky misalignment of the terms was performed before the subtraction. Let's do the proof again but let's replace 0.999... with the geometric series that it represents, namely 9/10 + 9/100 + 9/1000 + ... Then if we multiply it by 10 we get 90/10 + 90/100 + 90/1000 + ... then if we subtract what we started with we get 81/10 + 81/100 + 81/1000 + ... then if we divide by 9 we get back to exactly what we started with. We can easily show that a 'shift-and-subtract' operation on geometric series can result in contradiction. We start with 1+1+1+... = 1+1+1+... and simply subtract what we started with except that on one side we 'shift the terms' before doing the subtraction. We end up with 0=1 and so this appears to be evidence of why this 'proof that 0.999...=1' is flawed. There are other ways to demonstrate that we can't just re-arrange terms in a series. For example, we could re-write 0.999... as the series [9/10 + term1 - term1] + [9/100 + term2 - term2] + [9/1000 + term3 - term3] + ... where term1, term2 etc. are the terms of any conditionally convergent series. Then as per the Riemann Rearrangement Theorem we could make the result of the subtraction supposedly converge to any value we like. Most people agree that the base 10 decimal and the geometric series represent exactly the same thing and so should be interchangeable. And so if we can show there is a flaw in the proof when using geometric series, then the proof is no good at all. We can't claim the proof 'works' for the base 10 decimal but not for the geometric progression because these two forms are just different ways of writing exactly the same thing. As far as I can gather, most mathematicians think it is acceptable to define our way out of the problem. Mathematicians have simply invented a lot of rules about what we are allowed to do in terms of rearrangement/regrouping of terms. For example, we are not allowed to re-align the terms of a divergent geometric series and this allows us to claim that the example of 1+1+1+... = 1+1+1+... is not relevant. And (presumably) we are not allowed to modify an absolutely convergent series in any way where the result of a subtraction could be made to converge to more than one value. So even though we might think that the the example of 1+1+1+... = 1+1+1+... clearly demonstrates why the misalignment of endless non-zero terms in geometric series should not be allowed, it is not enough to simply work from 1st principles. We need to be aware of every definition and every rule that has ever been accepted into the discipline called 'mathematics' just in case there are any that say "that is not allowed". In other words, when faced with any counter argument to what mathematicians want to be true they say "that is not allowed". It is not a very elegant way to side-step problems to simply use a bunch of made-up rules that say "that is not allowed" and it raises the question of what exactly does mathematical rigour mean if mathematicians can simply define their way around any problem that they don't like.

This guy looks English

but as a conditionally convergent series, this sum is not unique. A rearrangement would lead to a different sum. Both the positive part and the negative part of this series are infinite in magnitude. According to Riemann's theorem, we can find a rearrangement that sums to ANY real number. So that one particular rearrangement is ln 2 is to be expected, that can be said of every conditionally convergent series.

Hi

On "speed", I turn off the sound, turn on the captions, set the speed to one quarter, and don't look at you gesturing. Plenty of time and fewer distractions. Highly recommend using screen videos. Work out your presentation on the computer, record as you go. Explain the steps and move the pieces around yourself in real time. It takes some preparation and you can do it by editing too. But I find talking heads mostly useless and distracting. You are putting your face between the people and the material. Metaphorically sit beside them and let them see the same screen that you do. Whisper in their ear or write your explanation. If the caption is well written the Google translate can convert the whole (taken is pieces because they have a size limit) caption listing and translate it to other languages. They are getting better and anything in the other languages is better than none. The chance of your spoken English to text and the translated to other languages -- coming out even close -- is small. You could do these same exercises, demonstrations, walk-throughs, verifications, checks, proofs, examples -- on a web page that supports symbolic mathematics with integrated simulations and visualizations (more than simple graphs). Numerics are not bad. You used that mentally when you said every term (1/(1+x)) would be smaller than the corresponding 1/x. But you could have showed the actual values. In real problems in engineering, sciences, economics, finance, budgeting, planning, project and device design and myriad other places, you will have to also do the quantitative demonstrations - for yourself and others. I am looking at all mathematics on the Internet - all of the tools, people, uses. This video presentation has to go into human eyes and minds. And in the minds find matching resonances (memories that are close). But with you in between, the people are not doing it themselves, except those who know how to do it by paper and pencil. Or rarely who have a symbolic math tool they can afford and is not so bloated it takes forever to learn or afford or use. Don't tell people how to fish, or show them pictures of fish and fishing - give them fishing poles, some hooks and bait, directions to where to fish, and give them a community of others doing the same thing. RU-vid only has this clunky blog thing. No index, no faqs, no collaboration, no building something together. There are horrific problems in the world, and people are only sharing words about the problems, not sharing the data and tools and organizations to map ans solve the problems. You know some of them. And truthfully, most of the posed mathematical problems only have value in the real world when the insights and more efficient methods help save lives, assure the survival of the human species, teach not hundreds or thousands - but billions of people who need better skills, better tools, and efficient project management for the really large, hard, but not impossible problems of the world. I am working on "covid" which has 7.5 billion entries on the Internet. That is mostly massive duplicates and variations on a core of a few hundred pages of knowledge and essential guidelines. Think about the problem of 4.8 Billion people who have some access to the Internet. They search for "covid" related things and get 7.5 Billions starting points and an almost infinite number of pathways. None of the material is traceable (much of it). Now you or I might spend hours or days or months tracing out tiny pieces, and sometimes we can find the author and tools and data and check things. But that level of diligence, and that amount of time and effort and skill - must be multiplied by the number of potential views (4.8 Billion) to get an estimate of the cost of our current messy Internet on society. "mathematics" on the Internet has 320 Million entry points. Many mouths shouting -- "I can tell you about mathematics". It is not good for society as a whole when all the mathematicians in the world are not working together - not to create a monopoly on tools and methods, jobs and applications - but to document what is known, what is being worked on, and share it - in the form of tools that new generations can use. I am working on "atomic fuels", "faster than light communication", "global climate change", "solar system colonization" and many thousands of similar topics. 7.8 Billion people is a lot of people and human time. The basic human brain is capable of much more than we ever require of it. And it can grow and stretch with good training data sets and tools for exploring and visualizing. You show symbolic mathematical steps. But your could have the computer do it, and let people step though the formal symbolic substitutions and operations to see how they are done. The processes of the steps are the product of some human learning of what is important to do when comparing quantities, structures and patterns. We should not make billions of people learn by rote listening to talking heads, but give them real tools and real data. dissemination If a paper is good, then millions might read it. But if it is poorly written, and it takes hours or days or weeks to understand just the pieces, then it is a burden on all those reader that has quantifiable economic and social impact. For covid, the delays mean 200,000 deaths each month. Just because the world is playing that game of "whisper" or "telephone" where a child whispers something in the ear of the next, and it gets repeated many times, - coming out garbled at the end. It always is garbled. For children at a birthday party it is a big laugh. But when millions of lives depend on something and the people involved all over the earth persist in sending "papers" to each other, rather than putting tools and data and models and active communities online -- it is just sad. Thank you, I like your skills and efforts, but I think you could also help with problems in the world. Training, not entertaining, people to use math for serious problems would help. Not memorizing steps, but finding and using good tools that can be used for global scale problems . Richard K Collins, Director, The Internet Foundation

early yayy

this is wrong

Please make the video for sum of 1-1+1-1+………………………

You said that a series is convergent because they approach 0 and each term is smaller than the previous is wrong. I find it ironic since you're talking about the alternating harmonic series while the normal harmonic series is divergent.

4:08