Have to show in $\mathbb{Q}_3$,
$$\sum_{n=1}^{\infty} \frac{3^{2n}(-1)^n}{4^{2n}n}=2\sum_{n=1}^{\infty} \frac{3^{2n}}{n4^n}$$
I subtracted above two things and tried to show but its not helping and can't think of any way either. Could anyone give a hint ?Qqua
Letting $$f(x)=\sum_{n=1}^{\infty} \frac{x^n}{n},$$ this is defined for $|x|_3<1$, and you get the identity:
$$f(x)+f(y)=f(x+y-xy)\tag{1}$$
Which is true because we have the identity in the complex numbers that: $f(x)=-\log(1-x)$, and so you have the power series identity: $$f(x)+f(y)=-\log((1-x)(1-y))=-\log(1-(x+y-xy))$$ In particular:
$$2f(x)=f(2x-x^2)$$
Now, try that with $x=-\frac{9}4$ so $2x-x^2=\frac{18}{4}-\frac{81}{16}=-\frac{9}{16}$.
The trick is that, although we initially know the identity (1) is true for the complex numbers with $|x|<1$, it is actually true as formal power series, and thus true for any topological field where the various series converge.