inequality question on real numbers #2 Expirimentally it seems that
$$\sum_{1\leq j\leq \lfloor n/8 \rfloor}\left(\frac{\pi en}{4j}\right )^{j/2}<2^{c_0n}$$
where $c_0=0.6$ and large $n$. Is there any proof?
Thank you.
 A: I don't think so.
Here are my calculations,
until gawk says the terms
are too large:
n = 100, log2(t1) = 38.6178, log2(t1)/n = 0.386178
n = 300, log2(t1) = 150.176, log2(t1)/n = 0.500585
n = 500, log2(t1) = 275.509, log2(t1)/n = 0.551019
n = 700, log2(t1) = 408.453, log2(t1)/n = 0.583505
n = 900, log2(t1) = 546.715, log2(t1)/n = 0.607461
n = 1100, log2(t1) = 689.073, log2(t1)/n = 0.62643
n = 1300, log2(t1) = 834.765, log2(t1)/n = 0.642127
n = 1500, log2(t1) = 983.268, log2(t1)/n = 0.655512
Also,
your expression is
$\sum_{j\leq n/8}\sqrt{\dfrac{\pi en}{4j}}^j$,
but to get results that seemed to match yours,
I had to write it as
$\sum_{j\leq n/8}\sqrt{\dfrac{\pi e(n^j)}{4j}}$.
A: Well I made the following computations in maple,
A=$log_2$the left part of the inequality and $B=c_0n$ then
n:=100, A=25.59,B=60
n:=200, A=51.88,B=120
n:=1000, A=256.61,B=600
n:=20000, A=5118.47, B=12000
so the expirimental results seems to confirm the inequality
A: Set $\displaystyle f_n(x)={\Big (}\frac{c_n}{x}{\Big)}^{x/2}\ (x>0), c_n=\pi\cdot e\cdot n/4$ and $g_n(x)=\log_2f_n(x).$
The extrema of $g_n$ is for $x=x_0=\pi\cdot n/4$ and 
$$g_n(x_0)=\frac{\pi\cdot n}{8\ln{2}}<0.57\cdot n.$$
So $f_n(x)<2^{0.57}$ and $$\sum_{1\leq j\leq n/8}f_n(x)<2^{0.57n}n/8<2^{0.6n}$$
for large enough $n.$
