If #a_k in RR^+# and #s = sum_(k=1)^na_k#. Prove that for any #n > 1# we have #prod_(k=1)^n(1+a_k) < sum_(k=0)^n s^k/(k!)#?

To prove the inequality \( \prod_{k=1}^{n}(1+a_k) < \sum_{k=0}^{n} \frac{s^k}{k!} \), we can use the inequality \( e^x > 1+x \) for all real \( x \). Let \( x = a_k \) for \( k = 1, 2, ..., n \). Then, we have \( e^{a_k} > 1 + a_k \) for each \( k \). Multiplying these inequalities for \( k = 1, 2, ..., n \), we get: \[ e^{a_1} \cdot e^{a_2} \cdot \ldots \cdot e^{a_n} > (1 + a_1)(1 + a_2) \cdot \ldots \cdot (1 + a_n) \] This simplifies to: \[ e^{a_1 + a_2 + \ldots + a_n} > \prod_{k=1}^{n} (1 + a_k) \] Since \( s = \sum_{k=1}^{n} a_k \), we have \( e^s > \prod_{k=1}^{n} (1 + a_k) \). Now, consider the power series expansion of \( e^s \): \[ e^s = \sum_{k=0}^{\infty} \frac{s^k}{k!} \] Truncating this series at \( n \), we obtain: \[ \sum_{k=0}^{n} \frac{s^k}{k!} \] Therefore, we have: \[ e^s > \prod_{k=1}^{n} (1 + a_k) \] \[ \sum_{k=0}^{\infty} \frac{s^k}{k!} > \prod_{k=1}^{n} (1 + a_k) \] \[ \sum_{k=0}^{n} \frac{s^k}{k!} > \prod_{k=1}^{n} (1 + a_k) \] Hence, we have proven that: \[ \prod_{k=1}^{n}(1+a_k) < \sum_{k=0}^{n} \frac{s^k}{k!} \]