Summing An Interesting Series

[@ExtinctTrain671]

In my real analysis textbook, this was about when power series were being thought of and Leonhard Euler came up with his Basel problem solution with the same concept. Pretty cool!!!

[@dan-florinchereches4892]

Nice approach Sybermath. But I still like Michael Penn's method in his video "A nice approach to the alternating harmonic series"

[@EternalSlumberer]

I asked this question to my math teacher in class hoping to kill some time, but he just uses the mc laurin expansion for ln(1+x)

[@Straight_Talk]

And so on and so fourth…

[@jahblohnsteron]

lol

[@abdisalanmohamed1564]

Thank u Teacher

[@holyshit922]

Looks like ln of 2 The other sum diverges

[@s1ng23m4n]

1 - x + x^2 - x^3 + ... = 1 + (-x) + (-x)^2 + (-x)^3 + ... = 1 / (1 - (-x)) = 1 / (1 + x)

[@elpescado2201]

I think i saw once that conmutating the terms of the series is only valid if you already know it converges... so, is there a way to show that it is convergent before actually computong the sum?

[@robertveith6383]

Is the following a hint? S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... S = (1 - 1/2) + (1/3 - 1/4) + ... S = 1/2 + 1/12 + ... So, S > 1/2. Also, S = 1 - (1/2 - 1/3) - (1/4 - 1/5) - ... S = 1 - 1/6 - 1/20 - ... So, S < 1. Therefore, 1/2 < S < 1.

[@pavlopanasiuk7297]

​@@robertveith6383 that doesn't speak of convergence. Alternating +1-1 series is also constrained, but it's divergent. The convergence here is guaranteed to Leibniz check. It is a much broader question on whether ln(1+x) would accurately describe the corresponding power series on its convergence boundary (which it does here, but it isn't guaranteed). Complex analysis, in my memory, doesn't speak about boundary convergence

[@mohamedomrane5481]

so long your method😯

[@Belgi_an_pizza]

You could have put r=-x and it'd work out😢

[@shmuelzehavi4940]

|x| mut be less than 1 otherwise, f'(x) does not converge to 1/(x+1).

[@Don-Ensley]

problem Can you sum 1 -1/2 + 1/3-1/4+1/5-1/6+... Let f(x) = ln( 1 + x ) We derive a Maclaurin series for ln( 1 + x ). Take the derivatives of f(x). f¹(x) = ( 1 + x )⁻¹ f²(x) =(-1) 1! ( 1 + x )⁻² f³(x) =(-1)² 2! ( 1 + x )⁻³ f⁴(x) =(-1)³ 3! ( 1 + x )⁻⁴ : : fⁿ(x) =(-1)⁽ⁿ⁻¹⁾ (n-1)! ( 1 + x )⁻ⁿ : : The Taylor series is f(x) = f(a) + f¹(a)(x-a)/1! + f²(a)(x-a)² /2! + f³(a)(x-a)³ /3! +... = ln( 1 + a )+ (1+a)⁻¹(x-a)/1! + (-1)1!(1+a)⁻²(x-a)² /2! + (-1)² 2! (1+a)⁻³ (x-a)³ /3! +... = ln(1+a)+ ͚ Σ(-1)⁽ⁿ⁻¹⁾(n-1)!(1+a)⁻ⁿ (x-a)ⁿ/(n!) ⁿ⁼¹ = ln(1+a)+ ͚ Σ(-1)⁽ⁿ⁻¹⁾(1+a)⁻ⁿ (x-a)ⁿ/n ⁿ⁼¹ Expand about a= 0 for the Maclaurin series. ͚ ln(1+x) = Σ(-1)⁽ⁿ⁻¹⁾ x ⁿ/n ⁿ⁼¹ Note that for x=1, this series is identical to the one we are finding. ͚ ln(2) = Σ(-1)⁽ⁿ⁻¹⁾/n ⁿ⁼¹ = 1 -1/2 + 1/3 -1/4 + 1/5-1/6+... The series ͚ S= Σ 1/n = 1 + 1/2 + 1/3 + 1/4 +... ⁿ⁼¹ is called the harmonic series, and is divergent. S → ∞ answer ln(2)