Power series at $x=0$ $$\sum_{n=0}^∞n!x^{2n}$$
This power series has radius of convergence $R=0$ since the ratio test shows it diverges for all $x≠0$.
But what if $x=0$? Then we have $\sum_{n=0}^\infty n! \cdot 0^{2n}$
Surely this series is undefined since the first partial sum is $0!\cdot 0^{2\cdot0}$ and $0^0$ is undefined, right?
Well, my lecturer claims this series is convergent and is equal to 1.
But $\sum_{n=1}^\infty n!\cdot 0^{2n}$ clearly converges to 0, so he is claiming that $0!\cdot 0^0=1$?
So is he wrong? Or am I missing something here?
 A: There are a lot of situations where it is convenient to define
$$0^0=1$$
and of course $0!=1$, so $0!\cdot0^0=1$.
In some situations, it's most appropriate to leave $0^0$ undefined. In particular, the function $x^y$ converges to different limits as $x\to0$ and $y\to0$ depending on which path is taken, so analysis does not yield a natural value for $0^0$ and it may not make sense to define $0^0=1$.
However, when dealing with discrete exponents (natural numbers), it's usually better to define $0^0=1$, because this is consistent with combinatorial or set-theoretic interpretations of exponentiation. For instance, the set of functions from the empty set to the empty set has cardinality $1$, so $0^0=1$ by the set-theoretic definition. Also, using the repeated multiplication definition of exponentiation, an empty product is always $1$, even when the product contains only zeroes. (For more information, see this Wikipedia article.)
It is important to consider where the $0^0$ in the expression comes from. In your case, it's the first term of a power series. A power series' notion of exponentiation is closest to the repeated multiplication interpretation, so taking $0^0$ makes sense. Consider that a power series is just a generalization of a polynomial. How would you evaluate $f(x)=x^2+x+1$ at $x=0$? What if this is written $f(x)=x^2+x^1+x^0$?
So your lecturer is probably correct.
