Big theta proof

**Sneaky** · January 20th 2013, 04:46 PM

Hi,

I need some help with this big theta proof:

Show that f(n) = summation (from h=1 to n) of h³/2^h 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 h³/2^h <= 1c₂
f(n) = 1/2 + 8/4 + ... + n³/2ⁿ <= c₂
1/2 <= c₂and 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...

**johng** · January 20th 2013, 06:51 PM

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

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

**Sneaky** · January 20th 2013, 07:06 PM

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?

**johng** · January 20th 2013, 07:31 PM

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

**Sneaky** · January 20th 2013, 07:44 PM

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

**johng** · January 20th 2013, 09:34 PM

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

**Sneaky** · January 21st 2013, 05:46 PM

nevermind, I get the question now. thanks

**Sneaky** · January 21st 2013, 07:38 PM

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

**johng** · January 23rd 2013, 09:00 AM

Yes and no; see the attachment:

$Big theta proof-mhflimitexplanation.png$

**Sneaky** · January 23rd 2013, 05:02 PM

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?