upper bound on sum of exponential functions Let $$\sum_{k=1}^{N} |a_k| = A$$ where $A$ is some constant. I am looking for an upper bound on the sum 
$$\sum_{k=1}^{N} e^{|a_k|},$$ independent of $N$, but may depend on $A$. That is, I am looking for something like
$$\sum_{k=1}^{N} e^{|a_k|} \le f(A).$$ 
We can find the upper bound
$$\sum_{k=1}^{N} e^{|a_k|} \le Ne^{A}.$$ 
but I do not want this dependence on N.
Does anybody have any ideas? From some reason I can't seem to figure this one out. Something like this
this would work, but I'm not sure its correct.
 A: The original problem is equivalent to maximizing the function
$$
E:[0,\infty)^N\to [0,\infty),\quad E(x)=\sum_{i=1}^Ne^{x_i},
$$
subject to
$$
K(x)=A,
$$
where
$$
K:[0,\infty)^N\to [0,\infty),\quad K(x)=\sum_{i=1}^Nx_i.
$$
If $a\in K^{-1}(A)$ is an extremum of $E$ on $K^{-1}(A)$, then there is a real number $t$ (this is called a Lagrange multiplier) such that
$$
\nabla E(a)=t\nabla K(a),
$$ 
i.e.
$$
\left(e^{a_1},\ldots,e^{a_N}\right)=t(1,\ldots,1).
$$
Thus $t=e^{a_1}=\ldots=e^{a_N}$, and consequently $a_1=a_2=\ldots=a_N$. Since $K(a)=A$, it follows that $a_i=\frac{A}{N}$ for all $i=1,2,\ldots,N$, and $E(a)=Ne^{\frac{A}{N}}$. Since $b=(0,\ldots,0,A)\in K^{-1}(A)$, and
$$
E(a)-E(b)=Ne^{A/N}-(N-1+e^A)=N(e^{A/N}-1)-(e^A-1)=\sum_{k=2}^\infty\left(\frac{1}{N^{k-1}}-1\right)\frac{A^k}{k!}\le 0,
$$
this shows that $a$ is not a point of maximum. Therefore, the maximum of our problem lies on the boundary $\partial K^{-1}(A)$.
The boundary of $K^{-1}(A)$ is
$$
\partial K^{-1}(A)=\bigcup_{m=1}^{N-1}\underbrace{\left[K^{-1}(A)\cap\left(\{0_{\mathbb{R}^m}\}\times\mathbb{R}^{N-m}\right)\right]}_{B_m}.
$$
Using the method of Lagrange multipliers, we show that $E$ has a maximum on $B_m$. This maximum is achieved at the point $b^{m} \in B_m$ such that 
$$
b_i^m=\begin{cases}
0 & \mbox{ for  } 1\le i\le m,\\ 
\frac{A}{N-m} &\mbox{ for } m+1\le i\le N
\end{cases}.
$$
Then
$$
E(b^m)=m+(N-m)e^{\frac{A}{N-m}},
$$
and for $n>m$ we have
\begin{eqnarray}
E(b^m)-E(b^n)&=&m+(N-m)e^{\frac{A}{N-m}}-n-(N-n)e^{\frac{A}{N-n}}=(N-m)\left[e^{\frac{A}{N-m}}-1\right]-(N-n)\left[e^{\frac{A}{N-n}}-1\right]\\
&=&\sum_{k=2}^\infty\left(\frac{1}{(N-m)^{k-1}}-\frac{1}{(N-n)^{k-1}}\right)\frac{A^k}{k!}\le 0.
\end{eqnarray}
Hence
$$
E(x)\le E(b^{N-1})=N-1+e^A \quad \forall x\in K^{-1}(A).
$$
A: There is no upper bound independent of $N$, as every term in the sum is at least $1$, so the sum is at least $N$.
A: Assume they are sorted,
so
$a_k \ge a_{k+1}
$.
Then
$\sum_{k=1}^n e^{a_k}
=e^{a_1}\sum_{k=1}^n e^{a_k-a_1}
\le n e^{a_1}
$
with equality iff
all the $a_k$ are equal.
This is pretty simple-minded,
but to do more
requires knowledge about the $a_k$.
