# 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!)?

Jun 29, 2017

Define "f"(n) = sum_(k=0)^n s^k/(k!).

"f"'(n)= sum_(k=0)^n ks^(k-1)/(k!),
"f"'(n) = sum_(k=0)^(n-1) s^k/(k!),
$\text{f"'(n) = "f} \left(n - 1\right)$.

$\text{f} \left(n\right)$ is increasing for $n \in {\mathbb{Z}}^{+}$ if $\text{f} ' \left(n\right) > 0$. $\text{f} ' \left(n\right) > 0$ if $\text{f} \left(n - 1\right) > 0$. $\text{f} \left(0\right) = 1$. Inductively, $\text{f} \left(n\right)$ is increasing for $n \in {\mathbb{Z}}^{+}$ and therefore $\text{f} \left(n\right) > 0$.

I'm going to prove a useful inequality.

Consider $\text{g} \left(x\right) = x - \ln \left(1 + x\right)$. Then $\text{g} ' \left(x\right) = 1 - \frac{1}{1 + x}$. Clearly $\frac{1}{1 + x} < 1$ for $x > 0$, so we conclude that $\text{g} \left(x\right)$ is an increasing function. As $\text{g} \left(0\right) = 0$, $\text{g} \left(x\right) > 0$ for $x > 0$. Then,

$\ln \left(1 + x\right) < x$

As $x > 0$ and ${a}_{k} > 0$, let $x = {a}_{k}$. We have,

$\ln \left(1 + {a}_{k}\right) < {a}_{k}$.

Summing each side,

${\sum}_{k = 1}^{n} \ln \left(1 + {a}_{k}\right) < {\sum}_{k = 1}^{n} {a}_{k}$.

We can substitute our definition of $s$.

${\sum}_{k = 1}^{n} \ln \left(1 + {a}_{k}\right) < s$.

By using $\ln \left(a\right) + \ln \left(b\right) = \ln \left(a b\right)$ and writing $s = \ln \left({e}^{s}\right)$ we conclude,

$\ln \left({\prod}_{k = 1}^{n} \left(1 + {a}_{k}\right)\right) < \ln \left({e}^{s}\right)$.

By taking an inverse logarithm,

${\prod}_{k = 1}^{n} \left(1 + {a}_{k}\right) < {e}^{s}$.

Define "h"(n) = sum_(n+1)^(\infty) s^k/(k!).

Then,

"h"'(n) = sum_(k=n+1)^(\infty) k * s^(k-1)/(k!),
"h"'(n) = sum_(k=n)^(\infty) s^k/(k!).
$\text{h"'(n) = "h} \left(n - 1\right)$.

Then $\text{h} \left(n\right) > 0$ if $\text{h} \left(n - 1\right) > 0$. Inductively, as $\text{h} \left(0\right) = {e}^{s}$ and ${e}^{s} > 0$, $\text{h} \left(n\right) > 0$.

So, we have that,

${\prod}_{k = 1}^{n} \left(1 + {a}_{k}\right) < {e}^{s}$,
prod_(k=1)^n (1+a_k) < sum_(k=1)^(n) s^k/(k!) - sum_(k=n+1)^(\infty) s^k/(k!),
prod_(k=1)^n (1+a_k) < sum_(k=1)^(n) s^k/(k!).