Inadvertently, I find this interesting inequality. But this problem have nice solution?

prove that

ln2>(25)25This problem have nice solution? Thank you.

ago,I find this

ln2<(12)12=√22following is my some nice methods,

use this inequality

x−ylnx−lny>√xy,x>y

then we let

x=2,y=1so ln2<√22

solution 2:

since

1n+1≤12⋅34⋯2n−12n

then

ln2=∞∑n=01(n+1)2n+1<∞∑n=0(2n)!(n!)223n+1=1√2solution 3

since

(1+√2)2(t+1)−(t+1+√2)2=t(1−t)>0

so

ln2=∫101t+1dt<∫10(1+√2t+1+√2)2dt=√22

solution 4:ln2=34−14∞∑n=11n(n+1)(2n+1)<34−14(11×2×3−12×3×5)=710<√22

solution 5

1√2−ln2=∞∑n=1√2(4n2−1)(17+2√2)n>0But ln2>(25)25 I can't have this nice solution

Thank you everyone can help.

**Answer**

This problem seems to be so hard to prove "elegantly" without aid of calculator, is because you have managed to find such a good approximation! One would need to estimate really well to be able to prove that one is greater than the other.

Continued fractions can be used to give a proof:

(25)2=425=16+14 is a convergent of the

the continued fraction of (log2)5=[0;6,4,592,1,…]

If you take the CF of (log2)5 for granted, the proof of what you want falls right out: 425 is a convergent which is smaller (the convergents alternate bigger/smaller).

Note the apperance of the huge 592 term, which tells you that 425 will be a good approximation, because the convergent corresponding to the 592 term is greater than (log2)5.

Perhaps a more compelling "reason" for it being a good approximation is that the continued fractions of log2=[0;1,2,3,1,6,3,1,…] and (25)2/5=[0;1,2,3,1,6,3,2,…], match up to a good 6 terms! The 7th term tells you that log2 must be greater: the parity of the position where two CFs first differ determines if the one with the greater number in that position is greater or not (which also explains the alternating property mentioned above).

If you absolutely want a proof which needs no calculator (i.e. can be verified manually in an hour or so :-)), here is one (with calculations missing):

Take the power series

log(1+x)−log(1−x)=2∞∑n=0x2n+12n+1

and set x=13, you get log2 on the left.

Now you can truncate the power series at any point, and get a smaller number than log(1+x)−log(1−x) , as coefficients are all positive.

Now if you truncate the series at n=4 (include n=4 term) you get the value 42976066200145 which is greater than (25)2/5. This you can see (manually) by computing the fifth power and computing the (numerator of the) difference of the two fractions you get.

I won't go into more details, as they are quite tedious to do so completely manually without the aid of any calculators, and this is probably not what you were expecting anyway.

Interestingly, it might be easier to do the computations if you work in base 9 or 3 (because of the powers of 13, you can quickly read off some of the digits, like spigot algorithms).

**Attribution***Source : Link , Question Author : math110 , Answer Author : Aryabhata*