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What is the upper bound of $\sum_{n=1}^{N}d_{n}2^{-\frac{b}{a_n}}$ where $a_n, d_n$ is non negative values?

Or can i find the solution $b$ satisfying following constraint ? $\sum_{n=1}^{N}d_{n}2^{-\frac{b}{a_n}}=C$

where $C$ is positive constant.

Using the following arithmetic-geometric inequality, I can find the lower bound. $\sum_{n=1}^{N}p_n x_n\geq\prod_{n=1}^{N}{x_n}^{p_n}$

But, I don't have any idea or mathematical theory to find upper bound.

Update: I modify the question.

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    Are there any other restrictions on the $a_n, b_n, d_n$? Otherwise the expression can be come arbitrarily large. – 2017-02-19
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    One more constraint is that $b_n=b$ for all $n$. And $b$ is non negative value. – 2017-02-19
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    Accordingly, the euqation can be rewritten as $\sum_{n=1}^{N}d_n2^{-\frac{b}{a_n}}$. My main consideration is how to integrate the form $2^{-\frac{b}{a_n}}$ for all $n$. If $a_n=a$ for all $n$, I can reformulate equation as $2^{-\frac{b}{a}}\sum_{n=1}^{N}d_n$. But, Because $a_n$ has different value for all $n$, I cannot integrate $2^{-\frac{b}{a_n}}$ . – 2017-02-20
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    Let $c_n=2^{-1/a_n}$ then the sum is $f(b)=\sum_{n=1}^N d_nc_n^{\,b}\,$. If any of the $c_n \gt 1$ then $f(b)$ is unbounded when $b \to \infty$. As for the edited question, the equation $f(b)=C$ does not have a closed form solution in the general case (though it may reduce to a polynomial equation if, for example, all $a_n$ are integers, but even in that case it doesn't have a closed form solution in general). – 2017-02-20
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    Dear, dxiv. Thnks for response. In my situation, always $0$a_n\geq1$. In this case, as you mentioned, the closed form is not exist. Accordingly, I want to know the upper bound form which can be tightly bounded. – 2017-02-20
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    I suspect there are many more hidden constraints which you have not yet added to the question. – 2017-02-20

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