# Function that Represents Divergent Power Series?

Suppose we have the following power series

$$\sum_{k=0}^\infty\left(x^2+1\right)^{2k}$$

If we wished to find the function that represents this series, it seems reasonable to suppose that the function $f$ that may represent this series is

$$f(x) = \frac{1}{1-\left(x^2+1\right)^2} = -\frac{1}{x^2(x^2+2)}$$

which is defined over $\mathbb{R^*}$

However, since we are assuming that the power series is geometric, the summand must be less than 1; therefore,

$$(x^2+1)^2<1$$

But this is absurd, since the $\min{x}=0$ renders the value 1. Therefore, the power series diverges.

Here is my question: Since the power series diverges, does this mean that it cannot represent the function that we deduced above for all $x$, in which case, the series does not define $f$?

• $\sum_{k=0}^\infty\left(x^2+1\right)^{2k}$ is a misused notation of a serie. We loosely accept to write it down to infinity for convenience purpose, but in fact we should always pay attention to the convergence first... In that case, this notation has no meaning. And thus, any "conclusion" you may have found by using this notation has no solid ground. – Martigan Jun 22 '15 at 7:29
• If you were to consider finite sums of terms, you could use the formula for the geometric series sum - this would have the extra term ((1+x^2)**2)**n+1, which would not vanish as n->infinity, so your step deducing the function f would be incorrect. – user247608 Jun 22 '15 at 8:03

The series has a well-defined real value for $x=\lambda\mathrm i$, where $-\sqrt2<\lambda<0$ or $0<\lambda<\sqrt2$ and $\mathrm i^2=-1.$ The algebraic formula you gave is valid for the analytic continuation of this function (on the imaginary interval pair $(-\sqrt2\:,0)\mathrm i\cup(0\;,\sqrt2)\mathrm i$) to all $x$ in the complex plane excluding the points $0$ and $\pm\sqrt2\mathrm i$, which includes the whole real line except $0$. This does not mean that the series defines a real function in the usual sense, by plugging in real numbers in place of $x$.