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$$\sum_{n=0}^∞n!x^{2n}$$

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?

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    $\begingroup$ $0^0 = 1$ by convention. $\endgroup$ – user258700 Dec 22 '15 at 2:01
  • $\begingroup$ Really? I've never heard that. I've always thought it was just undefined. Why isn't $\frac00$ similarly defined by convention? $\endgroup$ – Refnom95 Dec 22 '15 at 2:08
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    $\begingroup$ It's only when dealing with limits that $0^0$ is indeterminant. $\endgroup$ – Michael Burr Dec 22 '15 at 2:09
  • $\begingroup$ It's done mostly in the context of power series. It's much neater to write $\sum_i \text{bla}_ix^i$ than $\text{bla}_0+\sum_{i>0}...$. $\endgroup$ – YoTengoUnLCD Dec 22 '15 at 2:20
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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.

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