1. ## Big theta proof

Hi,

I need some help with this big theta proof:

Show that f(n) = summation (from h=1 to n) of h3/2h is big theta(1).

What I got so far:

omega(g(1)) <= f(n) <= O(g(1))
c1g(1) <= f(n) <= c2g(1)

For f(n) <= O(1)
f(n) = summation (from h=1 to n) of h3/2h <= 1c2
f(n) = 1/2 + 8/4 + ... + n3/2n <= c2
1/2 <= c2 and 8/4 <= c2 ... and n3/2n <= c2

Since n3/2n is the biggest term in the summation, then showing n3/2n <= c2 is enough to show that f(n) <= O(1).
And here is where I get stuck, I'm not sure what to do next...

2. ## Re: Big theta proof

Your assertion that $\displaystyle {h^3\over 2^h}$ is increasing is unfortunately false. In fact the given sequence is decreasing for h > 5 -- examine the function $\displaystyle f(x)={x^3\over 2^x}$.

Entirely new direction: does $\displaystyle \sum^\infty_{h=1}{h^3\over 2^h}$ converge? If so, does this mean your partial sums are $\displaystyle \Theta(1)$?

3. ## Re: Big theta proof

Oh ok, I see that when h > 5, it decreases. It then converges at 0 when h -> infinity.

But how can I evaluate the limit of the summation at h = infinity?

4. ## Re: Big theta proof

Let $\displaystyle S(n)=\sum^n_{h=1}{{h^3\over 2^h}$. Then to say the series converges is to say the limit as n approaches infinity of $\displaystyle S(n)$ is some finite value s -- you don't need to know specifically the value of s. So there is an $\displaystyle N$ such that for $\displaystyle {n\geq N}$, $\displaystyle s-1<S(n)<s+1$. So $\displaystyle c_1=s-1$ and $\displaystyle c_2=s+1$ satisfy the requirement of $\displaystyle \Theta$. Remember $\displaystyle \Theta$ requires the inequality only for all $\displaystyle n\geq N$; i.e. only for "large" values of n.

5. ## Re: Big theta proof

But for this, don't you need to show that limit is a finite number? How can this be done?

6. ## Re: Big theta proof

When we say a series converges, we mean $\displaystyle \lim_{n\rightarrow \infty}S(n)=s$, s a (finite) real number. Remember the ratio test for convergence: $\displaystyle \lim_{h\rightarrow\infty}{(h+1)^3/2^{h+1}\over h^3/2^h}=1/2<1$, so the series converges.

7. ## Re: Big theta proof

nevermind, I get the question now. thanks

8. ## Re: Big theta proof

shouldn't it be s-1 <= s(n) <= s+1 instead of s-1 < s(n) < s+1 ?

9. ## Re: Big theta proof

Yes and no; see the attachment:

10. ## Re: Big theta proof

I don't under stand the last two sentences.

If s(n) (- (s-1, s+1), how is it that s(n) (- [s-1, s+1] ?

Doesn't normal parenthesis like '(' and ')' mean, up to but not including?

11. ## Re: Big theta proof

Surely you know $\displaystyle 3<4$ implies $\displaystyle 3\leq4$. More generally, $\displaystyle (0,4)\subset[0,4]$; i.e. if $\displaystyle 0<x<4$ then $\displaystyle 0\leq x\leq4.$