I don't know how I should solve this exercise:
The polynomial ring $K[T]$, and hence also its field of fractions $K(T)$, is a subring of $K((T))$. Give an example, for some field $K$ , of an element $a \in K (( T )) $ with $ a \not\in K(T)$, such that is algebraic over $K ( T )$
I know that I have to search for a field and an element that is root of minimalpolynomial in that field but not an element of the field itself. For example $i \not\in \mathbb{R}$ but $f(x)=x^2 + 1 \in \mathbb{R}$ . But I don't know any example that is in the formal power series $K((T))$ but not in the field itself solving the algebraic condition. I'm thankful for help.
The way I read the question we are free to choose the field $K$. Just in case I misunderstood, I proffer an example for all fields. To that end it suffices to cover all the prime fields $\Bbb{Q}, \Bbb{F}_p$, $p$ a prime number.
Recall the binomial series of $\sqrt{1+T}$ (=its Taylor series centered at the origin) from calculus. Its coefficients are rational, so this works for any field $K$ of characteristic zero.
If $K$ has characteristic $p>2$, then you can first show that the coefficients of the series of $\sqrt{1+4T}$ are all integers, and thus make sense modulo $p$.
If $K$ has characteristic $p=2$, then apply freshman's dream to the series $$ a=T+T^2+T^4+T^8+\cdots=\sum_{k=0}^\infty T^{2^k} $$ to show that $a+a^2=T$. Mind you, this series has a counterpart for all $p$, so you can skip the integrality property in the previous suggestion, and use that instead.