Another problem by G.H. Hardy

Algebra Level 4

Define the function ϕ ( n ) = ( 1 + 1 n ) n \phi(n) = \left(1 + \dfrac1n\right)^n .

For n ≥ 1 n \geq 1 , which is greater, ϕ ( n ) \phi(n) or ϕ ( n + 1 ) \phi(n+1) ?

Note : A previous version of this problem did not have n ≥ 1 , n \geq 1, so those who answered "impossible to determine" and "it depends" have been given credit.

Impossible to determine They are equal ϕ ( n + 1 ) \phi(n+1) ϕ ( n ) \phi(n)

Hamza A by Hamza A , 19, USA

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2 solutions

Théo Leblanc
Jun 28, 2018

First write Φ(x)=exp(xln(x+1)-xln(x)) Thus to show Φ is an increasing function for x greater than one, it is suffisant to show that h: x-> xln(x+1)-xln(x) is an increase function for x greater than 1 because exp is an increasing function.

For x≥1, h'(x)= ln(x+1)-ln(x)-1/(x+1)= int(1/(1+t)dt from x-1 to x) - int(1/(1+x)dt from x-1 to x)

So h'(x)= int( (1/(1+t)-1/(1+x))dt from x-1 to x) And very obviously, g:x->1/(1+x) is a decreasing function for x≥1 so the inside part of the integral is always positive (and 0 for t=x) so the integral is positive, ie h'(x)>0.

So we can conclude that Φ is an increasing function on [1;+∞] and so Φ(n+1)>Φ(n)

Sorry if my English is not perfect, it is pretty hard :) Also I don't kown how to write integrals... If someone knows!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Sub in n=1. The answer is hence obtained.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

but phi(n) may be greater sometimes while it is less at other cases

Hamza A - 5 years, 4 months ago

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