You may have seen the harmonic series, but have you seen it in alternation?

Calculus Level 3

1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + … = ? \dfrac {1}{1} - \dfrac {1}{2} + \dfrac {1}{3} - \dfrac {1}{4} + \frac{1}{5} - \frac{ 1}{6} + \ldots = \ ?


The answer is 0.693.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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2 solutions

Discussions for this problem are now closed

Curtis Clement
Feb 26, 2015

Using the Taylor series: L n ( x + 1 ) = x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + . . . = ∑ n = 1 ∞ x n n ( − 1 ) n − 1 Ln(x+1) = x - \frac{1}{2}x^2 +\frac{1}{3}x^3 - \frac{1}{4}x^4 +... = \displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} (-1)^{n-1} ∴ ∑ n = 1 ∞ ( − 1 ) n − 1 n = L n ( 1 + 1 ) = L n 2 ≈ . 0.693 \therefore\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = Ln(1+1) = Ln 2 \approx. 0.693

M i s t a k e t o a v o i d : \large \ Mistake \ to \ avoid: 1 − 1 2 + 1 3 − 1 4 + . . . = ( 1 + 1 3 + 1 5 + . . . ) − ( 1 2 + 1 4 + 1 6 + . . . ) 1 - \frac{1}{2} +\frac{1}{3} - \frac{1}{4} +... = (1 + \frac{1}{3} + \frac{1}{5} +...) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} +...) = ( 1 + 1 2 + 1 3 + 1 4 + . . . ) − ( 1 + 1 2 + 1 3 + 1 4 + . . . ) = 0 = (1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+...) - (1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+...) = 0 T h i s i s n o t t r u e b e c a u s e : \large \ This \ is \ not \ true \ because: ( 1 + 1 2 + 1 3 + 1 4 + . . . ) − ( 1 + 1 2 + 1 3 + 1 4 + . . . ) = ∞ − ∞ ≠ 0 (1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+...) - (1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+...) = \infty - \infty \ne 0

22 Helpful 0 Interesting 0 Brilliant 0 Confused
Peter Macgregor
Feb 26, 2015

Because the harmonic series diverges, the given series is c o n d i t i o n a l l y conditionally convergent. This implies that (by a theorem due to Riemann) it can be rearranged to sum to a n y any number!

What we can prove is that provided − 1 < x < 1 -1<x<1

x 1 − x 2 2 + x 3 3 − x 4 4 + ⋯ = l n ( 1 + x ) \frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots=ln(1+x)

Proof

Given that − 1 < x < 1 -1<x<1 we may use the formula for a G.P. to write

1 − x + x 2 − x 3 + ⋯ = 1 1 + x 1-x+x^2-x^3+\dots=\frac{1}{1+x}

Integrating both sides from zero to x x then gives the above result.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

I like your proof of the Taylor series of L n ( x + 1 ) Ln(x+1) . Also, I would like to ask, why don't we know the expansion of L n ∣ x ∣ ? Ln|x|\ ?

Curtis Clement - 6 years, 3 months ago

I'll answer the second one. The first one is so long, I'm really lazy to write in.

For second problems you asked, we know that the Taylor expansion is: f ( x ) = ∑ n = 0 ∞ f n ( x ) ( x − a ) n n ! f(x)=\displaystyle\sum_{n=0}^{\infty} \frac{f^n(x) (x-a)^n}{n!} With: f k ( x ) = d k d x k f ( x ) f^k (x) = \frac{d^k}{dx^k} f(x)

If we put any real non-zero number of a a , it will look inefficient and sometimes, lazy to remember since it is too hard. When we substitute a = 0 a=0 , we get a Maclaurin series. Definitely, we can't expand ln ⁡ ( x ) \ln(x) since ln ⁡ ( 0 ) \ln(0) is − ∞ -\infty (sometimes undefined)

For the derivatives of ln ⁡ ( x ) \ln(x) let's take the 1st derivative:

d d x ln ⁡ ( x ) = 1 x \frac{d}{dx} \ln(x) = \frac{1}{x}

Substitute x = 0 x=0 , we get indeterminated result. Thus, we don't have any expansion for ln ⁡ ( x ) \ln(x) with a = 0 a=0

Why we don't know? To me, I'm too lazy to memorise too much and tedious formula. It's trivial. I don't mean to provoke, but my suggestion is we need to find a very efficient formula so we can easily understand and memorise a formula

These are my point of views.

Figel Ilham - 6 years, 3 months ago

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