A calculus problem by 敬全 钟

Calculus Level 4

1 1 + 1 2 − 1 3 − 1 4 + 1 5 + 1 6 − 1 7 − 1 8 + ⋯ \large \dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}+\cdots

If the closed form of the sum above can be expressed as π + ln ⁡ ( a ) a \dfrac{\pi+\ln (a)}{a} , where a a is a positive integer, find a a .


The answer is 4.

敬全 钟 by 敬全 钟 , 22, Malaysia

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3 solutions

This is essentially the composition of two series that is, ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 = π 4 \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} = \dfrac{\pi} {4} which is the series for tan ⁡ − 1 x \tan^{-1} x .

The other one is :

∑ n = 1 ∞ ( − 1 ) n − 1 n = ln ⁡ 2 \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}=\ln 2

Adding the first and half of the second we get the desired series and the result is π + ln ⁡ 4 4 \dfrac{\pi+\ln 4}{4} which makes the answer 4 \boxed{4}

8 Helpful 0 Interesting 0 Brilliant 0 Confused
Kushal Bose
May 19, 2017

The above expression can be written as

∑ r = 0 ∞ ( 1 4 r − 3 + 1 4 r − 2 − 1 4 r − 1 − 1 4 r ) \displaystyle \sum_{r=0}^{\infty} (\dfrac{1}{4r-3}+\dfrac{1}{4r-2}-\dfrac{1}{4r-1}-\dfrac{1}{4r})

= ∑ r = 0 ∞ ∫ 0 1 ( x 4 r − 4 + x 4 r − 3 + x 4 r − 2 + x 4 r − 1 ) d x =\displaystyle \sum_{r=0}^{\infty} \int_{0}^{1} (x^{4r-4}+x^{4r-3}+x^{4r-2}+x^{4r-1})dx

= ∫ 0 1 ( ∑ r = 0 ∞ ( x 4 r − 4 + x 4 r − 3 − x 4 r − 2 − x 4 r − 1 ) ) d x =\displaystyle \int_{0}^{1} (\sum_{r=0}^{\infty} (x^{4r-4}+x^{4r-3}-x^{4r-2}-x^{4r-1}))dx

= ∫ 0 1 ( 1 1 − x 4 + x 1 − x 4 − x 2 1 − x 4 − x 3 1 − x 4 ) d x =\displaystyle \int_{0}^{1} (\dfrac{1}{1-x^4} + \dfrac{x}{1-x^4} - \dfrac{x^2}{1-x^4}-\dfrac{x^3}{1-x^4}) dx

= ∫ 0 1 1 + x 1 + x 2 d x = π + ln ⁡ ( 4 ) 4 =\displaystyle \int_{0}^{1} \dfrac{1+x}{1+x^2} dx \\ =\dfrac{\pi+\ln (4)}{4}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
May 12, 2017

Consider Maclaurin series of the following:

− ln ⁡ ( 1 − x ) = x + x 2 2 + x 3 3 + x 4 4 + + x 5 5 + x 6 6 + x 7 7 + x 8 8 ⋯ Putting x = i − ln ⁡ ( 1 − i ) = i + i 2 2 + i 3 3 + i 4 4 + i 5 5 + i 6 6 + i 7 7 + i 8 8 + ⋯ Dividing both sides by i − ln ⁡ ( 1 − i ) i = 1 + i 2 + i 2 3 + i 3 4 + i 4 5 + i 5 6 + i 6 7 + i 7 8 + ⋯ i ln ⁡ ( 1 − i ) = 1 + i 2 − 1 3 − i 4 + 1 5 + i 6 − 1 7 − i 8 + ⋯ \begin{aligned} - \ln (1-x) & = x + \frac {x^2}2 + \frac {x^3}3 + \frac {x^4}4 + + \frac {x^5}5 + \frac {x^6}6 + \frac {x^7}7 + \frac {x^8}8 \cdots & \small \color{#3D99F6} \text{Putting }x = i \\ - \ln (1-i) & = i + \frac {i^2}2 + \frac {i^3}3 + \frac {i^4}4 + \frac {i^5}5 + \frac {i^6}6 + \frac {i^7}7 + \frac {i^8}8 + \cdots & \small \color{#3D99F6} \text{Dividing both sides by }i \\ - \frac {\ln (1-i)}i & = 1 + \frac i2 + \frac {i^2}3 + \frac {i^3}4 + \frac {i^4}5 + \frac {i^5}6 + \frac {i^6}7 + \frac {i^7}8 + \cdots \\ i\ln (1-i) & = 1 + \frac i2 - \frac 13 - \frac i4 + \frac 15 + \frac i6 - \frac 17 - \frac i8 + \cdots \end{aligned}

Therefore, we have:

S = 1 + 1 2 − 1 3 − 1 4 + 1 5 + 1 6 − 1 7 − 1 8 + ⋯ = ℜ [ i ln ⁡ ( 1 − i ) ] + ℑ [ i ln ⁡ ( 1 − i ) ] = ℜ [ i ln ⁡ ( 2 e − π 4 i ) ] + ℑ [ i ln ⁡ ( 2 e − π 4 i ) ] = ℜ [ i ( ln ⁡ 2 2 − π 4 i ) ] + ℑ [ i ( ln ⁡ 2 2 − π 4 i ) ] = ℜ [ π 4 + ln ⁡ 2 2 i ] + ℑ [ π 4 + ln ⁡ 2 2 i ] = π 4 + ln ⁡ 2 2 = π 4 + ln ⁡ 4 4 \begin{aligned} S & = 1 + \frac 12 - \frac 13 - \frac 14 + \frac 15 + \frac 16 - \frac 17 - \frac 18 + \cdots \\ & = \Re [i\ln (1-i)] + \Im [i\ln (1-i)] \\ & = \Re \left[i \ln \left(\sqrt 2 e^{-\frac \pi 4i} \right) \right] + \Im \left[i \ln \left(\sqrt 2 e^{-\frac \pi 4i} \right) \right] \\ & = \Re \left[i \left(\frac {\ln 2}2 -\frac \pi 4i \right) \right] + \Im \left[i \left(\frac {\ln 2}2 -\frac \pi 4i \right) \right] \\ & = \Re \left[\frac \pi 4 + \frac {\ln 2}2i \right] + \Im \left[\frac \pi 4 + \frac {\ln 2}2i \right] \\ & = \frac \pi 4 + \frac {\ln 2}2 = \frac \pi 4 + \frac {\ln 4}4 \end{aligned}

⟹ a = 4 \implies a = \boxed{4}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

For the first line, if -1<x <=1 is true, then x=i does not fall under this range.

Pi Han Goh - 4 years ago

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I was trying to resolve this but don't know how.

Chew-Seong Cheong - 4 years ago

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The issue is here is that you started with the geometric progression sum , 1 1 − x = 1 + x + x 2 + ⋯ \frac1{1-x} = 1 + x + x^2 + \cdots for − 1 < x < 1 -1<x< 1 (and then you integrate it).

After you integrate it, the domain of the function ∫ 0 x 1 1 − v d v \displaystyle \int_0^x \frac1{1-v} \, dv will still remain − 1 < x < 1 -1<x<1 (and might work at the endpoints as well, which can be proven via squeeze theorem ). So you can't say "Oh, I substitute x = i x = i and it just magically works!". To justify that x = i x= i works, you need to show that ∣ i ∣ = 1 |i|= 1 is at the end point of − 1 < x < 1 -1<x<1 , so you got to prove that − ln ⁡ ( 1 − x ) = ( what you have written up ) -\ln(1 - x) = (\text{what you have written up}) also converges when x = i x = i first.

I didn't solve it your way, but what I would have done is to convert all the fractions to the form of ∫ 0 1 x n d x \int_0^1 x^n \, dx , then summing over all the integrals. You should get Aditya's method.

Addendum:

Note that 1 − i = 2 ( 1 2 − i 2 ) = 2 cis ( − π 4 + 2 π k ) 1 - i = \sqrt 2 \left(\dfrac1{\sqrt2} - \dfrac{i}{\sqrt2} \right) = \sqrt 2 \text{cis}\left( -\dfrac\pi4 + 2\pi k \right) for any integer k k . So there are actually infinitely many values of ln ⁡ ( 1 − i ) \ln(1 - i ) . So, how can you justify that the final value you found is the unique answer?

Pi Han Goh - 4 years ago

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@Pi Han Goh – Good information.

Chew-Seong Cheong - 4 years ago

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