prove that $\frac{d (\ln F(t))}{dt}=\sum_{k>0}{(-1)^{k+1}(x_1^k+...+x_n^k)}t^{k-1}$

Question

$F(t)=(1+x_1t)(1+x_2t)...(1+x_nt)$, prove that $\frac{d (\ln F(t))}{dt}=\sum_{k>0}{(-1)^{k+1}(x_1^k+...+x_n^k)}t^{k-1}$.

Let's $n=1$, $F(t)=1+x_1t$, than $\frac{d (\ln F(t))}{dt}= \frac {x_1}{1+x_1t}$, $\sum_{k>0}{(-1)^{k+1}(x_1^k+...+x_n^k)}t^{k-1}=x_1$

so the main problem is that I don't understand how to deal with the denominator

Answer 1

You did a mistake in the second part of your calculation, which might help you understand better the problem:

$$\sum_{k>0}{(-1)^{k+1}(x_1^k+...+x_n^k)}t^{k-1}= x_1 \sum_{k > 0}{(-x_1 t)^{k-1}} $$

And if you know the sum of a geometrical series, you can see that this term then becomes $ \frac{x_1}{1+ x_1 t}$, which gives you exactly what you want!

The matter is now only to separate the sum for each $x_i$ and see that it works.