# How to find more precise estimation for fraction: $\frac{q^2 c1}{1+0.5q^2(1+c2)}$

I have a sum of fractions:

$$\sum\limits_{i=1}^n \frac{t^2\sigma_i^{2}(1-a^{\delta'})}{1+t^2\sigma_i^2(1+2a^{\delta'}(\frac{\cosh{\mu} - 0.5\mu^2 - 1}{\mu^2}))}$$

where $\delta' = \{1,2\}$, $1 \leq\mu \leq 2$, $t = \frac{\mu x}{B_n}$, $B_n^2 = \sum\limits_{i=1}^{n} \sigma_i^{2}$

and $x$ is big enough and $a$ is small enough.

I omit all $\sigma_i^2$ and suppose them like $B_n^2$ in denominator.

After that I estimate the denominator like this:

$e^x\geq (1+\frac{x}{y})^y$ for $x\gt 0 , y \gt 0$

And I came to this lower estimation for the sum above:

$$\frac{q^2c_1}{e^{(q(1+c_2))^{0.5}}},$$

where $q = x\mu$, $c_1 = (1 - a^{\delta'})$, $c_2 = 2a^{\delta'}(\frac{\cosh \mu - 0.5 \mu^2 - 1}{\mu^2})$

I need a more precise estimation, because this sum is an argument for exponential function.

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