
Congruence
Can somebody help with this question please?
Prove that
(e^x)(e^x2/2)(e^x^3/3)...=1+x+x^2+... when x<1.
Show that the coefficient of x^19 in the power series expansion on the LHS has the form
1/19! + 1/19 + r/s,
where 19 does not divide s.
Deduce that 18!= 1(mod 19).

Hi
$\displaystyle \prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}}$
To find $\displaystyle \sum_{k=1}^{+\infty}\frac{x^k}{k}$ let's take the derivative
$\displaystyle \sum_{k=0}^{+\infty}x^k = \frac{1}{1x}$
Therefore $\displaystyle \sum_{k=1}^{+\infty}\frac{x^k}{k}$ is the antiderivative of $\displaystyle \frac{1}{1x}$ that is 0 for x=0
$\displaystyle \sum_{k=1}^{+\infty}\frac{x^k}{k} = \ln1x$
Therefore $\displaystyle \prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}} = e^{\ln1x} = \frac{1}{1x}$

Thanks.
Is anybody able to help with the second part of the question please?

$\displaystyle e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots$
$\displaystyle e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{8\cdot 3!} + \cdots + \frac{x^{2n}}{2^n\cdot n!} + \cdots$
$\displaystyle e^{\frac{x^3}{3}} = 1 + \frac{x^3}{3} + \frac{x^6}{18} + \frac{x^9}{27\cdot 3!} + \cdots + \frac{x^{3n}}{3^n\cdot n!} + \cdots$
etc
To find the coefficient of x^19 in the power series expansion you need to multiply all these identities and find the coefficient of x^19

surely there must be an easier way to do this?

I hope so (Happy)
If you consider each term of the expansion of e^x from x^19 down to 1
 x^19/19! must be multiplied by a constant which is 1 therefore 1/19!
 x^18/18! must be multiplied by something times x, but this is not possible since in the other series are involved terms starting from x², therefore 0
 x^17/17! must be multiplied by something times x², which is x²/2 coming from the expansion of e^(x²/2) therefore 1/(2x17!)
 x^16/16! must be multiplied by something times x^3, which is x^3/3 coming from the expansion of e^(x^3/3) therefore 1/(3x16!)
 x^15/15! must be multiplied by something times x^4, which is x^4/4 coming from the expansion of e^(x^4/4) but also x^4/8 coming from the expansion of e^(x^2/2) therefore (1/4+1/8)/15! = 3/(8x15!)
 x^14/14! must be multiplied by something times x^5, which is x^5/5 coming from the expansion of e^(x^5/5) but also x^5/6 coming from the product of the expansion of e^(x^2/2) and the expansion of e^(x^3/3) therefore (1/5+1/6)/14! = 11/(30x14!)
and so on ... but it becomes more and more difficult (Wondering)

You guys are so close...
Yes, this is the ticket. Runninggag has the winning strategy, but don't think for a minute you have to find the actual coefficient. All the problem calls for is showing that it is of the form $\displaystyle \frac{1}{19!}+\frac{1}{19}+\frac{r}{s}$ , for 19 not dividing s.
For reference, call the series expansion $\displaystyle e^{\frac{x^k}{k}}$ by $\displaystyle a_k$, and the coefficient preceding $\displaystyle x^{kn} $ by $\displaystyle a_k(n)$ . For example, $\displaystyle a_1(3)=\frac{1}{3!}$ and $\displaystyle a_3(2)=\frac{1}{18}$ .
Now the coefficient on $\displaystyle x^{19}$ is the sum of all possible products of terms whose exponents sum to $\displaystyle 19$, for example, one of which is $\displaystyle a_1(4)*a_2(3)*a_3(3) = \frac{1}{4!}*\frac{1}{2^3 3!}*\frac{1}{3^3 3!}$
(i) $\displaystyle a_1(19)=\frac{1}{19!}$
(ii) $\displaystyle a_{19}(1)=\frac{1}{19}$
(iii) Since $\displaystyle 19$ is prime, nowhere else in any coefficient $\displaystyle a_k(n)$ does $\displaystyle 19$ appear, for $\displaystyle k<19$ and $\displaystyle n<19$.
Therefore, the sum of all such terms must be of the form $\displaystyle \frac{1}{19!}+\frac{1}{19}+\frac{r}{s}$ , for 19 not dividing s.
QED
EDIT: The coefficient is actually $\displaystyle \frac{1}{19!}+\frac{1}{19}+\frac{31546534213457683 52}{15687942664565435561256}$

Thanks
How do we deduce that 18! = 1 (mod 19) from all this?