An amazing summation!

Calculus Level 1

1 1 ⋅ 2 + 1 3 ⋅ 4 + 1 5 ⋅ 6 + 1 7 ⋅ 8 + … \large\dfrac1{1\cdot2}+\dfrac1{3\cdot4}+\dfrac1{5\cdot6}+\dfrac1{7\cdot8}+\ldots

Let S S denote the value of series above. Find the value of e S e^S .


The answer is 2.

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Ikkyu San by Ikkyu San , 23, Thailand

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2 solutions

Chew-Seong Cheong
Jun 23, 2015

S = 1 1 ˙ 2 + 1 3 ˙ 4 + 1 5 ˙ 6 + 1 7 ˙ 8 + . . . = 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + . . . = x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + 1 5 x 5 − 1 6 x 6 + . . . [ For x = 1 ] = ln ⁡ ( 1 + x ) = ln ⁡ ( 1 + 1 ) = ln ⁡ 2 ⇒ e S = e ln ⁡ 2 = 2 \begin{aligned} S & = \frac{1}{1\dot{}2} + \frac{1}{3\dot{}4} + \frac{1}{5\dot{}6} + \frac{1}{7\dot{}8} + ... \\ & = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + ... \\ & = \color{#3D99F6}{x} - \frac{1}{2}\color{#3D99F6}{x^2} + \frac{1}{3}\color{#3D99F6}{x^3} - \frac{1}{4}\color{#3D99F6}{x^4} + \frac{1}{5}\color{#3D99F6}{x^5} - \frac{1}{6}\color{#3D99F6}{x^6} + ... \quad \quad \color{#3D99F6}{[\text{For }x = 1]} \\ & = \ln{(1+\color{#3D99F6}{x})} = \ln{(1+\color{#3D99F6}{1})} = \ln{\color{#3D99F6}{2}} \\ \\ \Rightarrow e^S & = e^{\ln{2}} = \boxed{2} \end{aligned}

30 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Great! Can you find the flaw in the working below?

1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + … = ( 1 1 + 1 3 + 1 5 + 1 7 + … ) − ( 1 2 + 1 4 + 1 6 + … ) > ( 1 1 + 1 3 + 1 5 + 1 7 + … ) > ( 1 2 + 1 4 + 1 6 + 1 8 + … ) = 1 2 ( 1 1 + 1 2 + 1 3 + 1 4 + … ) = ∞ \begin{aligned} &&\frac11 -\frac12 +\frac13-\frac14+\frac15-\frac16+\ldots \\ &=& \left(\frac11+\frac13+\frac15+\frac17+\ldots \right) - \left(\frac12+\frac14+\frac16+\ldots \right) \\ &>& \left(\frac11+\frac13+\frac15+\frac17+\ldots \right) \\ & >& \left(\frac12+\frac14+\frac16+\frac18+\ldots \right) \\ &= &\frac12 \left(\frac11 + \frac12 + \frac13 + \frac14 +\ldots\right) = \infty \end{aligned}

@Calvin Lin The > > in the third line is incorrect. Besides, you can't rearrange a conditionally convergent series.

Jake Lai - 5 years, 11 months ago

A beautiful use of colors in the answer.

Sanjeet Raria - 5 years, 11 months ago

@Calvin Lin Even more Flawful! (1 + 1/3 + 1/5 + .... ) = s1 (1/2 + 1/4 + 1/6 + ....) = s2 s = (1 + 1/3 + 1/5 + .... )- (1/2 + 1/4 + 1/6 + ....) = s1 - s2

s2 = 1/2(1 + 1/2 + 1/3 + .....) = 1/2 ( (1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ....) ) = 1/2(s1 + s2) s2 = (s1 + s2 )/ 2 ???? s2 = s1?

Siva Bathula - 4 years, 2 months ago
Anuj Mishra
Jun 24, 2015

S = 1 1.2 + 1 3.4 + 1 5.6 + … S = \frac{1}{1.2} + \frac{1}{3.4} + \frac{1}{5.6} + \ldots

= ( 1 1 − 1 2 ) + ( 1 3 − 1 4 ) + … = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + \ldots

E x p a n s i o n f o r l n ( 1 + x ) = x − x 2 2 + x 3 3 − … Expansion \ for \ ln(1+x) = x-\frac{x^2}{2} +\frac{x^3}{3} -\ldots [ F o r ∣ x ∣ < 1 ] [For |x|<1]

For x → 1 , x \rightarrow 1,

We can a p p r o x i m a t e l y approximately say that

l n ( 1 + x ) = 1 − 1 2 2 + 1 3 3 − … = S ln(1+x) = 1 - \frac{1^2}{2} + \frac{1^3}{3} - \ldots = S

⇒ e S = e ln ⁡ ( 2 ) = 2 \Rightarrow \ e^S \ = e^{\ln(2)} = \boxed 2

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually it's convergent for x=1, no "approximately" needed. The behavior on the boundary of the radius of convergence is nontrivial but it turns out that only x=-1 diverges.

Vilim Lendvaj - 3 years, 1 month ago

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