Summing A Series In Two Ways? 😁

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Опубликовано:

18 сен 2024

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Комментарии : 30

@MrGeorge1896 3 дня назад

Or as @blackpenredpen calls it: Our best friend in math is 1 / (1 - x) = 1 + x + x² + ... 😀

@SyberMath 3 дня назад

😁

@tanmaygoel4142 3 дня назад

I have the easiest solution. As we know the expansion of ln(1+x) = x - x²/2 + x³/3 - x⁴/4 +...... Put x=1 ln2 = 1 -1/2 + 1/3 -1/4 + ....... Hence ln2 is the answer 😊

@SyberMath 3 дня назад

How do we know ln(1+x) = x - x²/2 + x³/3 - x⁴/4 +......?

@maxhagenauer24 2 дня назад

@@SyberMath It's the maclaurin series polynomial expansion of it. You can derive it using the taylor series formula.

@ShaunakDesaiPiano 2 дня назад

@@maxhagenauer24I think what he means is if we should be allowed to assume the Maclaurin series expansion is true.

@maxhagenauer24 2 дня назад

@@ShaunakDesaiPiano Which it is, I can say the same things about all the assumptions and formulas he used in his method.

@ianfowler9340 2 дня назад

@@SyberMath Hidden in your question is whether or not the series converges for x = 1.The Radius of Convergence for ln(x+1) is the open interval (-1,1). It says nothing at all about x = 1 or x = -1. You have to check the endpoints individually. You can't just sustitute in x = 1 without justification. How do you show that the Interval of Convergence is the interval (-1,1]? It's a legitimate question. I don't know the answer - yet.

@Gunslinger-us1ek 3 дня назад

wow this is amazing

@conrad5342 2 дня назад

I am surprised the sum works for a geometric series with the factor -1 too ... the beauty of the convergence radius. Sometimes it converges on it, sometimes it doesn't.

@ianfowler9340 2 дня назад

Concerning the Maclaurin Series for ln(x+1): I'm leaving out the details, but the Ratio Test gives the Radius of Convergence to be the open interval (-1,1). But this says nothing at all about what happens at x = 1 or x=-1. In general it may, or may not converge at none, only1, or both of the end points points. So do we have to check to see what happens at the endpoints separately? In this case it does converge when x = 1 (ln2) but not x = -1. So how do we prove that it does actually converge at x = 1? Maybe the answer is related to this: All infinite alternating series will converge if the limit of the terms of the series is 0. So : (a) 1 - 1/2 + 1/3 - 1/4 + 1/5 is clearly alternating (b) lim (-1)^(n+1)/n as n----> inf is equal to 0.

@bobbyheffley4955 День назад

Alternating harmonic series

@scottleung9587 3 дня назад

Incredible!

@ianfowler9340 3 дня назад

When you start with the infinite geom. series formula, convergence only occurs when -1

@fakenullie 2 дня назад

I guess you could argue that it's improper integral, so you evaluate it at Lim x->1-, but when it's -1 you get infinity

@snejpu2508 3 дня назад

I used the second method of course. But one thing is interesting. Take the equation x=x/2. The standard way to solve it would be multiplying both sides by 2, subtracting x from both sides, and there you go, x=0. On the other hand, if both series approach infinity and they are equal, technically the infinity satisfies the equation, as infinity/2 is infinity. But not really, as f(x)=x and g(x)=x/2 both approach infinity, however with other slopes, so they cannot really meet. Moreover, negative infinity would satisfy the equation as well... Anyway for x=x/2, if x is not equal to 0, you can divide both sides by x and get 1=1/2, which is a contradiction, so the assumption on x not eqaul to 0 is wrong, so x has to be equal to 0. :-)

@ramenguy6827 3 дня назад

Can you tell me how we know the series is not converging and it is diverging? Why it can't be 0

@chaosredefined3834 2 дня назад

@@ramenguy6827 An infinite sum's value is defined as the limit of the partial sums. So, in this case, 1, 1-1/2, 1-1/2+1/3, 1-1/2+1/3-1/4, etc... So, for it to converge to zero, we need it to get arbitrarily close to 0. Specifically, for any value e, we should be able to find some point in that series of partial sums such that all the points after it add up to something between -e and e. So, let's say e = 1/2. Let's look at an even term in the series after that point. So, we have 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) - 1/2n < 1/2. Well, 1 - 1/2 = 1/2. And, for any positive integer k, 1/(2k-1) > 1/2k, so 1/(2k-1) - 1/(2k) > 0. So, 1/3 - 1/4 > 0, 1/5 - 1/6 > 0, ... 1/(2n-1) - 1/(2n) > 0. So, we have (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + ... + (1/(2n-1) - 1/(2n)) = 1/2 + positive thing + positive thing + ... + positive thing > 1/2. So, the even versions of the partial sums are always greater than 1/2. Thus, the limit of the sequence can be, at minimum, 1/2. Thus, it can't be 0.

@giuseppemalaguti435 2 дня назад

Se ricordo bene è ln2

@Qermaq 3 дня назад

And 1 - 2 + 3 - 4 + 5 - 6 + ... = 1/4. :D

@terryendicott2939 3 дня назад

I like your proof that ln(2) =0.

@राजनगोंगल 3 дня назад

👍👍👍👍👍👍

@wonghonkongjames4495 День назад

Good Morning, Sir Sorry to interrupt，but I don't think the 1st method is acceptable as though it seems However, the reasoning behind it isn't gonna be revealed in our daily maths The second one is correct, I must say - although there may be some other ways to get the same result, too, I think Thank you, James 09-17-2024

@SyberMath День назад

Yes, you are right

@rakenzarnsworld2 3 дня назад

Answer: 1

@chaosredefined3834 2 дня назад

Incorrect. This is a famous series, it converges to ln(2).

@yusufdenli9363 3 дня назад

So, first solution is wrong??

@Pascal-cy9sc 3 дня назад

条件収束の場合は、項を入れ換えて足したり引いたりは、収束値が変わってしまうので、無限の世界では、項の入れ換えは成り立たない

@bsmith6276 2 дня назад

At 5:27 Sybermath brings up Absolute Convergence. That is why the first method fails to give a real answer. Rearranging the terms of a conditionally convergent series (not absolutely convergent) can give you any possible answer you want. In this case Sybermath rearranged stuff to make zero the answer.

@chaosredefined3834 2 дня назад

To make it clear that it's wrong... Note that 1 > 1/2, 1/3 > 1/4, etc... So, 1 - 1/2 > 0, 1/3 - 1/4 > 0, etc... So, 1 - 1/2 + 1/3 - 1/4 + ... > 0. That should have indicated that something was wrong.