I want to prove this directly
$$2^{\frac{1}{n+1}}

Question

I want to prove this directly
$$2^{\frac{1}{n+1}}<n+1~~for~~n\ge 1$$
I proved it by induction, i want to see a direct proof.

misty1212 · Answer

HI!!!

misty1212 · Answer

you could try showing that 
$$\frac{2^{\frac{1}{n+1}}}{n+1}<1$$

xapproachesinfinity · Answer

i actually tried something similar with logs

misty1212 · Answer

besides the fact that it is completely obvious right?

misty1212 · Answer

the limit is zero for sure

xapproachesinfinity · Answer

well it is, but that's not enough

xapproachesinfinity · Answer

analytically yes!

misty1212 · Answer

you want something more complicated? 
it is true if $n=1$ now show the function is decreasing

xapproachesinfinity · Answer

i showed it with calculus 
i don't want to use calculus somehow haha
i showed it is decreasing

xapproachesinfinity · Answer

direct proof i meant no calculus assumed. just elementary tools :)

rational · Answer

$$2^{\frac{1}{n+1}}<n+1~~for~~n\ge 1$$

is same as showing

$$2<2^n~~for~~n\ge 2$$

thomas5267 · Answer

$$ 2^{\frac{1}{n+1}}=2^{\frac{n+2}{(n+1)(n+2)}}>2^\frac{n+1}{(n+1)(n+2)}=2^{\frac{1}{n+2}}\text{ for }n\geq1\ 2^{\frac{1}{n+1}}>2^{\frac{1}{n+2}}>2^{\frac{1}{n+3}}>2^{\frac{1}{n+4}}>\cdots \ \text{But:}\ n+1<(n+1)+1<(n+2)+1<\cdots\ \text{So:}\ \cdots<2^\frac{1}{5}<2^\frac{1}{4}<2^\frac{1}{3}<\underbrace{2^\frac{1}{2}<2}_{n=1}<3<4<\cdots $$