Question

show that \[\large \sqrt[n]{n}\lt 2\] for all \(n\gt 0\)

Answer 1

is there a neat solution without product log function

Answer 2

@xapproachesinfinity claims to be having a neat solution.. he used this result in some other problem..

Answer 3

cool question, good night guys :)

Answer 4

try doing it by mathematical induction

Answer 5

\[\sqrt[n]{n}<2\]
as n>0, take left n right to the nth power will give
n < 2^n
which should be true for all n>0

am i missing something here? :)

Answer 6

thats slick xD

Answer 7

The question
n^(1/n) < 2

seems algebraically equivalent to, after raising both sides to power of n,
n < 2^n

Now can you show that 2^n > n for all n greater than or equal to 0 ?

Answer 8

@sdfgsdfgs

Answer 9

it can be shown by induction i think
2^1 > 1 , basis case for n=1
Assume : 2^k > k for some k >= 1
2^(k+1) = 2 * 2^k > 2 * k > k

Answer 10

\(\large\color{white}{.}\)

Answer 11

i think induction works if "n" is an integer

Answer 12

*positive integer

Answer 13

To show it is true for x continuous you can use calculus.
CLaim :
2^x > x , for x > 0
:equivalent to:
2^x - x > 0 for x > 0.

The derivative of the left side is positive for x > 0 , so it is increasing and remains greater than zero

Answer 14

did you not mean to use positive integers n, i assumed that since you wrote n > 0

Answer 15

positive real numbers

since we're trying for rigor, how easy is it to show that \(2^x\ln(2)-1\) is always positive compared to showing that \(2^x\gt x\) ?

Answer 16

2^x ln 2 -1 > 0
\( \iff \)
2^x ln 2 > 1
\( \iff \)
2^x > 1 / ln 2
\( \iff \)
2^x > 1.44
\( \iff \) if x = 1
2^1 = 2 > 1.44
and 2^x is an increasing function , which you can show by taking the derivative

Answer 17

the derivative isn't really positive for all x>0 right ?

Answer 18

2^x has a positive derivative for all x ,

Answer 19

i thought out function is 2^x-x ?

Answer 20

I wanted to demonstrate that
2^x - x is increasing function for x > 0 , and it is bounded below by 1

Answer 21

2^x-x is increasing for sure for x > 1
but it has a local minimum in interval (0, 1)

Answer 22

i want to prove that 2^x - x > 0 for all x > 0
graphically it is true for all x. but no matter

2^0 - 0 = 1 > 0

Answer 23

hmm, you have a point there, it does decrease a little at first, then it increases

Answer 24

ok another way to prove it is use binomial series

Answer 25

2^x = (1 + 1)^x < x

Answer 26

okay so your proof using derivatives actually works for x > 1

Answer 27

yeah

Answer 28

it works for x > c such that

2^c * ln 2 - 1 = 0

Answer 29

yeah nice
so are we using extended binomial theorem where the exponent can be a real number ?

Answer 30

thats actually a fun problem, solve that equation :)

Answer 31

( 1 + 1)^x = 1^x + x*1 + x(x-1)/2 * 1 + ...

Answer 32

\[2^x=e^{x\ln 2 } = 1+x\ln 2+\frac{x^2}{2}\ln^2 2+\cdots \]

Answer 33

just using the power series of e^x

Answer 34

http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_and_connection_to_the_binomial_series

Answer 35

does that prove it is bigger than x ?

Answer 36

hmm idk lol i don't see yet

Answer 37

graphically 2^x - x > 0 for all reals , surely there must be a way to prove this analytically , for all reals

Answer 38

yeah i like the different methods you're trying

Answer 39

it is decreasing on (-oo, c) = ln(1/ln(2)) / ln (2)
increasing on (c, oo)

Answer 40

f(x) = 2^x -x is decreasing on (-oo, ln(1/ln(2)) / ln (2) )
increasing on ( ln(1/ln(2))/ ln (2) , oo)

Answer 41

sorry...took a shower ;)

"show that 2^n > n for all n greater than or equal to 0 ? " where n is a positive real number (instead of a positive integer)

since both functions, 2^n and n, are continuous, their respective values over positive real number range are bounded by the values over positive integers.

Since n goes up linearly while 2^n goes up exponentially, the case is proven for all positive real numbers >1.

For n between 0 and 1, 2^n will always be >1 and hence >n.

So 2^n > n for all n >0 where n is a positive real no (n cannot be equal to 0 in the original inequality.)

Answer 42

I would try this
Sqrt(n)(1+1+.....+1 n times ) <2

Answer 43

i think its enough to show that it reaches the minimum at a positive value , and is decreasing before it, increasing after

Answer 44

So
Sqrt(n)(1+1+1+1..)< n sqrt(n)(1)

Answer 45

i dont know what theorem to call this approach

Answer 46

Ahh that works!
Also @sdfgsdfgs 's argument uses a simple observation : In the interval (0,1) we have 2^x > 1 and x < 1. Together imply 2^x > x.

looks neat to me

Answer 47

how do we prove that 2^x > 1 in (0,1)

Answer 48

haha good q, calculus i guess

Answer 49

2^0=1, 2^1=2 n 2^n is continuous between (0,1) so 2^1 must be between 1 and 2.

Answer 50

show that
2^x - 1 > 0
take derivative of 2^x -1
2^x * ln 2 -1 > 0
2^x > 1/ln 2

Answer 51

2^0=1, 2^1=2 and 2^n is continuous between (0,1) so 2^1 must be between 1 and 2.

Answer 52

Is there anyway to show that sqrt(n)(n) is not rational, I mean its clear its not integer for n>1 but testing rationality seems cool to me

Answer 53

that looks interesting

Answer 54

wil try a bit on the proof and post a new question :)

Answer 55

2^0=1, 2^1=2 and 2^x is continuous between (0,1)
`and 2^x increasing as its first derivative is always positive.`

That means 2^x cannot go below 1 in the interval (0,1).
so 2^x must be between 1 and 2.

Answer 56

since the derivative of 2^x is always positive, on the interval (0,1)
2^x > 2^0 = 1> x

then for x >= 1 , you can use the argument i used earlier
2^x - x > 0
take derivative of f(x) = 2^x - x
this is increasing for x >=1
and f(1) = 2^1 -1 = 1 >0

therefore 2^x - x >= 1 > 0 for x >= 1

therefore 2^x > x .

i hope this is complete proof

Answer 57

looks good to me !

Answer 58

maybe like this also can work:
let the value of this rooth be a, then \(a^n=n\), now taking natural logs, for example,
\(n\ln a= \ln n\), or \(\ln a= (\ln n)/ n\)
now, getting a back:
\(a=e^{(\ln n)/n}\)
maybe taking ln expanssion you could figure it out from there

Answer 59

i still think binomial series could prove enlightening here
2^x = (1+1)^x > x
or the Bernoulli inequality

Answer 60

Also can one prove from the power series expansion that
$$\large 2^x=e^{x\ln 2 } = 1+x\ln 2+\frac{x^2}{2}\ln^2 2+\cdots > x $$
This is a true statement for all x

Answer 61

wow you guys really took it further:)

Answer 62

Haha yeah I was thinking its more simple like this :_
2^n=2*...*2>=2+...+2=2n>n

Answer 63

But was having fun reading all comments that my solution sounds a bit boring :p