Suppose I have an (integer valued, but it probably does not matter) *monotonically increasing* function $f: \mathbb{N} \to \mathbb{R},$ which satisfies 
$$\sum\_{i=1}^n f(i) < f(2 n).$$ What can we say about the speed of growth of this function?



lower bounds for growth of a sequence
Here is the trivial part where I'm pretty sure that the way is close to the most elegant requested:


$f(n)=e^{c\log^2 n}$ with $c=\frac 1{2\log 2}$ satisfies the inequality.
Indeed, 
$$
f(2n)=e^{c\log^2 n+2c\log 2\log n+c\log^22}\ge f(n)n\ge \sum\_{i=1}^nf(i)
$$
Here is the crude lower bound $f(n)\ge \delta f(1)e^{c\log^2 n}$ with any $c<\frac 1{2\log 2}$ and $\delta=\delta(c)>0$.


For any finite number of terms the estimate holds if we choose $\delta>0$ small enough. Suppose that we have the estimate up to $2n-1$ with sufficiently large $n$. Then
$$
f(2n)\ge \delta\sum\_{0\le j<\varepsilon n}e^{c\log^2(n-j)}\ge\varepsilon ne^{c\log^2 n+2c\log(1-\varepsilon)\log n}
\\
=e^{c\log^2 n+\log n(1+2c\log(1-\varepsilon))+\log\varepsilon}\ge e^{c\log^2(2n+1)}
$$
provided that $c<\frac 1{2\log 2}$, $\varepsilon<\varepsilon(c)$, and $n\ge N(c,\varepsilon)$,
so we are fine up to $2n+1$.


I wouldn't call this "elegant" and it is way too crude (we can actually get an asymptotic behavior; then there will be a correction logarithmic term in the exponent, etc.), but I'll leave it here for now.


***Elementary upper bound for $p(n)$***.


First, choose how many different numbers you want in your partition (can choose any $k\le \sqrt{2n}$, so $\sqrt{2n}$ choices here.


Next choose the actual numbers you want to use. They should sum up to $\le n$, so choose $k$ points between $1$ and $n$ and view your numbers as the differences between successive points (including 0); if two of those coincide, discard the configuration, otherwise note that each set appeared $k!$ times, so we have ${n\choose k}/k!\le \frac{n^k}{(k!)^2}$ options here.


Now, choose the multiplicities so that the sum is $\le n$. Obviously, the worst case scenario is when you have $1,2,\dots,k$. If your students don't mind elementary geometry, just associate with each integer solution of $a\_1+2a\_2+\dots+ka\_k$ a parallelepiped in $\mathbb R^k$ of size $1\times 2\times\dots\times k$ with the corner at $(a\_1,2a\_2,\dots, ka\_k)$. Since $1+2+\dots +k\le n$, they are all contained in the simplex $x\_1+\dots+x\_k\le 2n$, $x\_j\ge 0$, so the volumetric bound gives $\frac{(2n)^k}{(k!)^2}$.


Now all we need to show is that $\frac{n^k}{(k!)^2}=\left[\frac{(\sqrt n)^k}{k!}\right]^2\le e^{2\sqrt n}$.


If your students are afraid of $k$-dimensional geometry or ignorant of the series expansion of $e^x$, let me know and I'll try to circumvent both :-)


