# Prove that 1+x+(x^2)/2!+...+(x^n)/n! <(or equal to) e^x

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### 1 Answer

We know that:

f(1) = 1+1/1!+1/2!+1/3!+1/4! +...........+1/n!+......

Let f(x) = 1+x/1!+x^2/3+x^3/!+......x^n/n!+........

Now the ration r+1th to and rth term, as r -->infinity Ur+1/Ur = x/r = 0 for all x.

Therefore, the series f(x) is convergent.

f(m) = 1+m/1!+m^2/2!+m^3/3!+......+x^m/m! +......

f(n) = 1+n/1!+n^2/2!+n^3/3!+.....+x^m/n+.......

f(m)*f(n) = 1+(m+n)/1!+{m^2/2!+mn/(1!*1!)+n^2!}+(m^3/3!+m^2*n/(2!*1!)+mn^2/(1!*2!)+m^3/3!}+....+(m^r/r!+m^(r-1)*n/((r-1)!*1!)+m^(r-1)*n^2/((r-2)!*2!+m^(r-3)*r^3/((r-3)!*3!)+...... n^r/r!}

=1+(m+n)/1!+(m+n)^2/2!+(m+n)^3/3!+.....(m+n)^r/r!+........

= f(m)*f(m).

Therefore, f(m+n) = f(m)*f(m) or

f(m)(f(n) = f(m+n)

So, f(m)f(m)f(p)..... = f(m+n+p+...).

Similarly, [f(1)}^x = f(1+1+1+...x times) = f(x), where x is integer.

If xis a positive fraction, p/q, where p and q are integers, then,

[f(p/q)]^q = f(p/q+p/q+p/q+....n times) = f(p) = [f(1)]^p = e^p or f(p/q) = e^(P/q) or

f(x0 = e^x.

Similarly we can prove f(x) = e^x, is true when x is -ve also. Thus e^x= 1+x/1!+x^2/2!+x^3/3!+...x^n/n!+x^(n+1)/(n+1)!+..... or

e^x >= 1+x/1!+x^2/2!+x^3/3!+.....+x^n/n!