An application of law of large numbers How can one apply a law of large numbers to a Poisson Process in order to deduce the analytic fact that 
$$\lim_{t\rightarrow\infty} e^{-t}\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}\left|1-\frac{t}{2n+1}\right|=0?$$
It seems at least that one would have to use a triangular array weak law since the Poisson Process is continuous time and the limit is continuous time, but I'm not even sure I see it then?

If $X_t$ is a Poisson random variable with parameter $t$ and $Y_t=1/(1+X_t)$, this is $$E(|1-tY_t|;X_t\ \text{even})\leqslant E(|1-tY_t|)\leqslant(E((1-tY_t)^2))^{1/2},$$ hence  it suffices to show that $E((1-tY_t)^2)\to0.$
Note that $$t\,E(Y_t)=t\mathrm e^{-t}\sum_{n\geqslant0}\frac{t^n}{n!}\frac1{n+1}=1-\mathrm e^{-t},\qquad t^2\,E(Y_t^2)=\mathrm e^{-t}s_t,\qquad s_t=\sum_{n\geqslant1}\frac{t^{n+1}}{n!\cdot n},$$ hence $$E((1-tY_t)^2)=2\mathrm e^{-t}-1+\mathrm e^{-t}s_t,$$ and the result follows if $s_t\sim\mathrm e^t$ when $t\to\infty$. In one direction, $$s_t\geqslant\sum_{n\geqslant1}\frac{t^{n+1}}{(n+1)!}=\mathrm e^t-1-t\sim\mathrm e^t.$$ In the other direction, for every $n\geqslant N$, $n\geqslant N(n+1)/(N+1)$ hence $$s_t\leqslant\sum_{n=1}^{N-1}\frac{t^{n+1}}{n!\cdot n}+\frac{N+1}N\sum_{n\geqslant N}\frac{t^{n+1}}{(n+1)!}=P_N(t)+\frac{N+1}N(\mathrm e^t-Q_N(t)),$$ for two polynomial functions $P_N$ and $Q_N$ of degree $N-1$. Can you conclude?
