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!)#,
#"f"'(n) = "f"(n-1)#.
#"f"(n)# is increasing for #n in ZZ^+# if #"f"'(n)>0#. #"f"'(n)>0# if #"f"(n-1)>0#. #"f"(0)=1#. Inductively, #"f"(n)# is increasing for #n in ZZ^+# and therefore #"f"(n)>0#.
I'm going to prove a useful inequality.
Consider #"g"(x) = x-ln(1+x)#. Then #"g"'(x) = 1 - 1/(1+x)#. Clearly #1/(1+x) < 1# for #x>0#, so we conclude that #"g"(x)# is an increasing function. As #"g"(0) = 0#, #"g"(x)>0# for #x>0#. Then,
#ln(1+x)<x#
As #x>0# and #a_k>0#, let #x=a_k#. We have,
#ln(1+a_k)<a_k#.
Summing each side,
#sum_(k=1)^(n) ln(1+a_k) < sum_(k=1)^n a_k#.
We can substitute our definition of #s#.
#sum_(k=1)^n ln(1+a_k) < s#.
By using #ln(a)+ln(b)=ln(ab)# and writing #s=ln(e^s)# we conclude,
#ln(prod_(k=1)^n (1+a_k)) < ln(e^s)#.
By taking an inverse logarithm,
#prod_(k=1)^n (1+a_k) < 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!)#.
#"h"'(n) = "h"(n-1)#.
Then #"h"(n)>0# if #"h"(n-1)>0#. Inductively, as #"h"(0)=e^s# and #e^s>0#, #"h"(n)>0#.
So, we have that,
#prod_(k=1)^n (1+a_k) < 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!)#.
