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Here is my proof of the following problem.

Problem. Let $c_{j,n}, j= 1, \dots n$ such that $$ \begin{cases} \max_{1 \leq j \leq n}|c_{j,n}| \to 0,\\ \qquad\displaystyle\sum_{j=1}^{n} c_{j,n} \to \lambda,\\ \sup_{n} \displaystyle\sum_{j=1}^{n} |c_{j,n}| < \infty \end{cases}\text{ as }n \to \infty. $$ Prove that $$\prod_{j=1}^{n}(1+c_{j,n}) \to e^{\lambda} \;\;\text{ as }n \to \infty.$$ My proof. We have $\max_{1 \leq j \leq n} |c_{j,n}| \to 0$, so $c_{j,n} \to 0 \implies \ln(1+c_{j,n}) \to c_{j,n}$

$$\implies\sum_{j=1}^{n} \ln(1+c_{j,n}) \to \sum_{j=1}^{n} c_{j,n}$$ as $n \to \infty$

$$\implies \prod_{j=1}^{n}(1+c_{j,n}) \to e^{\lambda}$$

I don't know my proof is correct or not because I don't understand the role of the last condition - $ \sup_{n} \sum_{j=1}^{n} |c_{j,n}| < \infty $. PLease help me check it and explain the role of this condition.

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  • $\begingroup$ Estimate $|\log(1+c_{j,n})-c_{j,n}|$ by $c_{j,n}^2$ and note that $\sum_j c_{j,n}^2 \le \lambda \times \max_j |c_{j,n}|$. This should be moved to Math.SE, though. $\endgroup$ – Mateusz Kwaśnicki May 23 at 9:42
  • $\begingroup$ First, clarify the statement $\ln(1+c_{j,n}\to c_{j,n}$. Be very very precise about what it means. Next, you need to clarify the statement $$\sum_{j=1}^n \ln(1+c_{j,n})\to \sum_{j=1}^n c_{j,n}.$$ Then you will understand why the condition $\sup_n \sum_{j=1}^n|c_{j,n}|<\infty$ is useful. $\endgroup$ – Liviu Nicolaescu May 23 at 10:08