Given: $$\max_{1\leq j \leq n} |c_{j,n}| \to 0$$ $$\sum_{j=1}^n c_{j,n} \to \lambda$$ $$\sup_n \sum_{j=1}^n |c_{j,n}| < \infty,$$ show that: $$ \prod_{j=1}^n (1 + c_{j,n}) \to e^\lambda.$$
My partial solution: Taking the log of both sides, the conclusion is equivalent to $$ \sum_{j=1}^n \ln (1 + c_{j,n}) \to \lambda.$$ Indeed, $$ \sum_{j=1}^n \ln (1 + c_{j,n}) = \sum_{j=1}^n \frac{\ln (1 + c_{j,n})}{c_{j,n}} c_{j,n}.$$ Let $$ \overline{c}_{j,n} := \max_{1\leq j \leq n} c_{j,n},$$ $$ \underline{c}_{j,n} := \min_{1\leq j \leq n} c_{j,n}.$$ Now define $$k_n := \frac{\ln (1 + \underline{c}_{j,n})}{\underline{c}_{j,n}},$$ and $$l_n := \frac{\ln (1 + \overline{c}_{j,n})}{\overline{c}_{j,n}}.$$ Clearly, for all $n$ and all $1 \leq j \leq n$, $$ l_n \leq \frac{\ln (1 + c_{j,n})}{c_{j,n}} \leq k_n.$$ Using the first assumption, $l_n,k_n \to 1$. Assuming that $\lambda>0$, we write $$ l_n \sum_{j=1}^n c_{j,n} \leq \sum_{j=1}^n \frac{\ln (1 + c_{j,n})}{c_{j,n}} c_{j,n} \leq k_n \sum_{j=1}^n c_{j,n},$$ and finally, using the squeeze theorem, and the second assumption, our result follows. (If $\lambda<0$, the inequalities switch sides.)
My question: I haven't used the last assumtion, so I'm guessing something is missing in my solution.
Your last inequality is not true. Indeed, $l_n\sum_n c_{j,n}$ may not be less than $\sum_n \log(1+c_{j,n})$ because $c_{j,n}$ can be negative. One way to prove this problem is by evaluating the difference.
$$ \left|\sum_{j=1}^n\log(1+c_{j,n})-\sum_{j=1}^nc_{j,n}\right|\le\sum_{j=1}^n\left|\frac{\log(1+c_{j,n})}{c_{j,n}}-1\right||c_{j,n}|\le \max_{1\le j\le n}\left|\frac{\log(1+c_{j,n})}{c_{j,n}}-1\right|\sum_{j=1}^n|c_{j,n}| $$
The last term goes to zero as you can check by the three assumptions.