Factorials are large.

Calculus Level 2

Is the following summation larger than 4?

1 ! 2 ! + 2 ! 3 ! + 3 ! 4 ! + … + 99 ! 100 ! \frac{ 1!} { 2!} + \frac{ 2!} { 3!} + \frac{ 3!} { 4!} + \ldots + \frac{ 99!} { 100!}

Yes Cannot be determined No

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Adam Phúc Nguyễn by Adam Phúc Nguyễn , 20, Vietnam

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5 solutions

Discussions for this problem are now closed

Jake Lai
Jan 9, 2015

First, we simplify each term. It is easy to see that ( n − 1 ) ! n ! = 1 n \frac{(n-1)!}{n!} = \frac{1}{n} , and so the sum becomes S = 1 2 + 1 3 + … + 1 100 S = \frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{100} .

This is one less than the 100th harmonic number H n H_{n} , which can be approximated by the asymptotic limit γ + ln ⁡ n \gamma+\ln n where γ \gamma is the Euler-Mascheroni constant . Taking n = 100 n = 100 and subtracting one gives

S ≈ γ + ln ⁡ 100 − 1 ≈ 4.182 > 4 S \approx \gamma+\ln 100 -1 \approx \boxed{4.182 > 4}

40 Helpful 0 Interesting 0 Brilliant 0 Confused

Please explain little more about the formula used ,thanks.

Krishna Garg - 6 years, 5 months ago

H n = 1 1 + 1 2 + 1 3 + … + 1 n − 1 + 1 n H_{n} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n} can be approximated by γ + ln ⁡ n \gamma+\ln n .

It turns out that γ = ∫ 1 ∞ 1 ⌊ x ⌋ − 1 x d x \displaystyle \gamma = \int_{1}^{\infty} \frac{1}{\lfloor x \rfloor}-\frac{1}{x} dx as well, along with satisdying other interesting equations like γ = − Γ ′ ( 1 ) = ∑ k = 1 ∞ ( 1 ⌊ k ⌋ 2 − 1 k − 1 k 2 ) \displaystyle \gamma = -\Gamma^{\prime}(1) = \sum_{k=1}^{\infty} \left( \frac{1}{\lfloor \sqrt{k} \rfloor^{2}}-\frac{1}{k}-\frac{1}{k^{2}} \right) .

Jake Lai - 6 years, 4 months ago

Wouldn't it be ln(n+1) - 1 (for the additional 1 added to make the series 1/n) and hence equal to 4.62-1 = 3.62?

Pratik Joshi - 6 years, 5 months ago

You forget that I've added the constant γ \gamma .

Jake Lai - 6 years, 4 months ago

I was thinking to make this an approximation to the integration of 1/x That answer yields 3.91 Seems that it is not that accurate

Koh Hui Soon - 6 years, 4 months ago

This is because you are taking I = ∫ 2 100 d x x I = \displaystyle \int_{2}^{100} \frac{dx}{x} rather than I ′ = ∫ 1 100 d x x I^{\prime} = \displaystyle \int_{1}^{100} \frac{dx}{x} , which is a careless mistake.

Reviewing the definition of Riemann sums will tell you why I ′ I^{\prime} is more accurate than I I .

Jake Lai - 6 years, 4 months ago
Roman Frago
Jan 15, 2015

The series is equivalent to 1 2 + 1 3 + 1 4 + . . . + 1 100 \frac {1} {2}+\frac {1} {3}+\frac {1} {4}+...+\frac {1} {100}

Let S be the sum S > 3 4 + 4 8 + 4 12 + . . . + 4 100 S>\frac {3} {4}+\frac {4} {8}+\frac {4} {12}+...+\frac {4} {100}

S > 3 4 + 4 ( 1 2 + 1 3 + 1 4 + . . . + 1 25 ) S>\frac {3} {4}+4(\frac {1} {2}+\frac {1} {3}+\frac {1} {4}+...+\frac {1} {25})

but ( 1 2 + 1 3 + 1 4 + . . . + 1 25 ) > 24 × 1 25 (\frac {1} {2}+\frac {1} {3}+\frac {1} {4}+...+\frac {1} {25})>24 \times \frac {1} {25}

S > 3 4 + 4 ( 24 25 ) S>\frac {3} {4}+4(\frac {24} {25})

S > 459 100 S>\frac {459} {100} which is greater than 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution has been marked wrong 1 2 ≯ 3 4 \frac {1}{2} \not > \frac {3 }{4} , so your second line is wrong.

Can you help me figure out why this approximation exceeds the asymptotic limit.

Roman Frago - 6 years, 4 months ago

I don't know how you derived the formula in the second line, however look at the 3rd line, on the RHS you have:

4 ( 1 2 + 1 3 + … 1 25 ) ≥ ( 1 2 + + 1 3 + … 1 96 ) 4\left(\frac{1}{2}+\frac{1}{3}+\dots\frac{1}{25}\right)\ge\left(\frac{1}{2}++\frac{1}{3}+\dots\frac{1}{96}\right) and also

3 4 ≥ 1 2 + 1 97 + 1 98 + 1 99 + 1 100 \frac{3}{4}\ge\frac{1}{2}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}

therefore you have to reverse the inequality sign

Barack Clinton - 6 years, 4 months ago
Tamoghna Purkait
Jan 15, 2015

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution has been marked incomplete because it fails to show the most crucial step.

Nax Alpha
Jan 10, 2015

This is in fact

∑ i = 2 100 i 2 \sum\limits_{i=2}^{100} i^2

And then Wolfram Alpha

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution has been marked wrong for obvious reasons. The summation clearly shows that it is the sum of squares of whole numbers, while the sum in question is actually the sum in reciprocal of whole numbers. Furthermore, the use of computational aid in not required in this question.

Are you sure?

Calvin Lin Staff - 6 years, 5 months ago

that is not like that?

Kamran Shabbir - 6 years, 5 months ago
Brock Brown
Jan 9, 2015 
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from math import factorial as fac
from fractions import Fraction as frac
total = 0
for n in xrange(1,100):
    total += frac(fac(n), fac(n+1))
print float(total)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

frac(1, n) is good enough :)

Hoai-Thu Vuong - 6 years, 5 months ago

ahh Python, very clever

Amy Morton - 6 years, 4 months ago

I think the range should be xrange(1,99)

Kamal Saleh - 6 years, 4 months ago

Not quite, xrange(1, 99) would output the result of the following:

1!÷2! + 2!÷3! ... 98!÷99! ≈ 4.177

This is because the parameters for the generator function xrange are start and stop . Start is the first number that xrange will output and stop is the number at which the generator function halts. The xrange function will never output stop .

I always thought that was weird, too, but just remember to make stop equal to 1 more than the number you want to end on.

Brock Brown - 6 years, 4 months ago

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