I have a series like this:$$\sum_{n=0}^∞\left(\frac{λδ_t}2\right)^n\frac1{n!}\sum_{j=0}^n(-1)^{n-j}C_n^j\exp\left(\frac{(n-2j)^2}{2δ^2}+\frac{(n-2j)γ}δ\right),$$where $C_n^j$ is the binominal coefficient, and $δ_t$ is the volatility of the GARCH model.
I tried to estimate the series $\displaystyle\sum_{j=0}^n(-1)^{n-j}C_n^j\exp\left(\frac{(n-2j)^2}{2\delta^2}+\frac{(n-2j)\gamma}{\delta}\right)$ like this: $$\exp\left(\frac{(n-2j)^2}{2\delta^2}+\frac{(n-2j)\gamma}{\delta}\right)\le \exp\left(\frac{n^2}{2\delta^2}+\frac{(n-j)\gamma}{\delta}\right),$$ so $$\sum_{j=0}^n(-1)^{n-j}C_n^j\exp\left(\frac{(n-2j)^2}{2\delta^2}+\frac{(n-2j)\gamma}{\delta}\right)\le \sum_{j=0}^n(-1)^{n-j}C_n^j\exp\left(\frac{n^2}{2\delta^2}+\frac{(n-j)\gamma}{\delta}\right)\\ =(-1)^n\exp\left(\frac{n^2}{2\delta^2}\right)\sum_{j=0}^n(-1)^{j}C_n^j\exp\left(\frac{(n-j)\gamma}{\delta}\right)=(-1)^n\exp\left(\frac{n^2}{2\delta^2}\right)(e^{\gamma/\delta}-1),$$ but since $\lim\limits_{n\to\infty}\dfrac{1}{n!}\exp\left(\dfrac{n^2}{2\delta^2}\right)=\infty$ the series does not converge.
Maybe there may be another upper bound which proves its convergence? Because the lower rude estimation $\exp\left(\dfrac{(n-2j)^2}{2\delta^2}+\dfrac{(n-2j)\gamma}{\delta}\right)\geqslant1$ gives a sum about $1$.