<s>[INST] Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like 
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me some hints/guides here?
Thanks.
 [/INST] Sure.
$$\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots$$
In particular, the terms alternate sign. If we subtract $x^2$, we are left with
$$\ln(1 + x^2) - x^2 = -\frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \ldots$$
The first term of the sum is negative, and since the sum is a strictly alternating sum whose terms go to $0$ (since $|x| < 1$), the error between a sum of the first $n$ terms and the actual value is bounded by the $n+1$st term. Here, we only need to note that the first term is negative and the second term is smaller in absolute value - thus we conclude that 
$$\ln(1 + x^2) - x^2 \leq 0$$
with equality only when $x = 0$.
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