Your assertion that is increasing is unfortunately false. In fact the given sequence is decreasing for h > 5 -- examine the function .
Entirely new direction: does converge? If so, does this mean your partial sums are ?
Hi,
I need some help with this big theta proof:
Show that f(n) = summation (from h=1 to n) of h^{3}/2^{h} is big theta(1).
What I got so far:
omega(g(1)) <= f(n) <= O(g(1))
c_{1}g(1) <= f(n) <= c_{2}g(1)
For f(n) <= O(1)
f(n) = summation (from h=1 to n) of h^{3}/2^{h} <= 1c_{2}
f(n) = 1/2 + 8/4 + ... + n^{3}/2^{n} <= c_{2}
1/2 <= c_{2 }and 8/4 <= c_{2 } ... and n^{3}/2^{n} <= c_{2 }
Since n^{3}/2^{n} is the biggest term in the summation, then showing n^{3}/2^{n} <= c_{2} is enough to show that f(n) <= O(1).
And here is where I get stuck, I'm not sure what to do next...
Let . Then to say the series converges is to say the limit as n approaches infinity of is some finite value s -- you don't need to know specifically the value of s. So there is an such that for , . So and satisfy the requirement of . Remember requires the inequality only for all ; i.e. only for "large" values of n.