Let $(X_n)_n$ be a sequence of independent random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x).$ Let $U_n=\sum_{k=1}^nX_k.$
Prove that $(U_n)_n$ converges in distribution if and only if $\sum_{n}\alpha_n\ln(1+\dfrac{1}{\lambda_n})<+\infty.$
Supposing that $(U_n)_n$ converges in distribution to $U$, this means that $E[e^{-U_n}]=\prod_{k=1}^n E[e^{-X_k}]=\prod_{k=1}^n(\frac{\lambda_k}{\lambda_k+1})^{\alpha_k}$ converges to $E[e^{-U}]>0$ so $\sum_n\alpha_n \ln(1+\frac{1}{\lambda_n})<+\infty.$
For the converse, I don't know if this attempt is true: $\sum_n\alpha_n \ln(1+\frac{1}{\lambda_n})<+\infty,$ which means $E[e^{-U_n}]$ converges to a limit $l>0$ so $P(\sum_nX_n<+\infty)>0$ so $P(\sum_n X_n<+\infty)=1$ since this event is tail event. We conclude since a.s converges implies convergence in distribution.