I know the proof using binomial expansion and then by monotone convergence theorem. But i want to collect some other proofs without using the binomial expansion.  *if you could provide the answer without the concept of calculus it will be appreciated * But any way do provide whatever method you can.
Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI.
Travel without an Italian residence permit in the Schengen area cgi and tomcat im trying to run a cgi script (.cgi) with tomcat. I am getting the below error and cant find out whats wrong. I know i should really use apache and mod proxy but this really isnt my area of expertise so im taking the easy way out!  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