Divisor Function t(n): Proving $2n*(n)^{1/2}$

[peteryellow]

Let t(n) be the divisor function, i.e., the function gives the positive divisors in n including 1 and n. Then I want to show that

$2n*(n)^{1/2}.$

I have tried different ideas but nothing is working can somebody please give some hints.

 

[HallsofIvy]

Let t(n) be the divisor function, i.e., the function gives the positive divisors in n including 1 and n. Then I want to show that

$2n*(n)^{1/2}.$

I have tried different ideas but nothing is working can somebody please give some hints.

I'm sorry, I don't understand what it is you want to prove. Is that last formula just $2nn^{1/2}= 2n^{3/2}$ and what does it have to do with t(n)?

 

[peteryellow]

soory .

t(n) = divisor function, i.e., number of positive divisors of n including n and 1.

Then I want to show that t(n) < 2*(n)^{1/2}. I mean 2 times squareroot of n.

 

[dodo]

If the prime factorization of n is p1^k1 . p2^k2 . p3^k3 . ..., then t(n) = (k1+1)(k2+2)(k3+3)... Using this, and the fact that the logarithm function is monotonic, maybe (just maybe) taking logs on both sides of the inequality might help. Just an idea.

 

[ialbrekht]

Dodo:

then t(n) = (k1+1)(k2+2)(k3+3)

You started in interesting direction, but $$f(n) = k!$$ where $$k=\sum(k_i)$$

Let's start iterative theorem prove:
k=1
then $$t(n)=1$$ and $$1<2\sqrt{n}$$ for any $$n>1$$.

Prove for k+1
$$t(n)=(k+1)!=k!(k+1)$$
Now we have to prove that $$k!(k+1) < 2\sqrt{n}$$
Let's suppose that it's not true. Divide both sides on k+1, than we'll have $$k! > \frac{2}{k+1}\sqrt{n}$$ (1). But we know that $$k!<2\sqrt{n}$$ and $$\frac{2}{k+1}\sqrt{n}<2\sqrt{n}$$.

This means that (1) is impossible and $$k!(k+1) < 2\sqrt{n}$$ is true.