Consider the following: we have a triangular array of nonnegative numbers $$x_{1,1 } \\x_{2,1 } \ x_{2,2 } \\ x_{3,1 } \ x_{3,1 } \ x_{3,3 } \\... $$
The maximum on each row converges to zero: $\max_{1 \le k \le n } x_{n,k }\to 0$, and the sequence of row sums converges to $\lambda > 0 $: $\sum _{k=1 } ^n x_{n,k } \to \lambda$ as $n \to \infty $.
Does it follow that $$\prod_{k=1 } ^n (1+x_{n,k } ) \to e^{\lambda }\ ?$$
My hunch is that for $n $ large $x_{n,k } $ shoud be close to $\frac {\lambda } {n } $ and thus $\prod_{k=1 } ^n (1+x_{n,k } ) \to e^{\lambda } $ is close to $(1+\frac {\lambda } {n })^n $ which approaches $e^{\lambda } $.
Thanks in advance!
Yes, it does. As $0\leqslant x-\ln(1+x)\leqslant x^2/2$ for $x\geqslant 0$, we have $$0\leqslant\sum_{k=1}^{n}\big(x_{n,k}-\ln(1+x_{n,k})\big)\leqslant\frac{1}{2}\sum_{k=1}^{n}x_{n,k}^2\leqslant\frac{1}{2}(\max_{1\leqslant k\leqslant n}x_{n,k})\sum_{k=1}^{n}x_{n,k}\underset{n\to\infty}{\longrightarrow}0$$ which implies $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\ln(1+x_{n,k})=\lim_{n\to\infty}\sum_{k=1}^{n}x_{n,k}=\lambda$ and your claim.