Taylor polynomial around $x=0$ for $\ln(2+x)$

Question

For my homework I got this question: explain what is wrong with this statement: if $f(x)=\ln(2+x)$, then the second-degree taylor polynomial approximating $f(x)$ around $x=0$ has a negative constant term. I proceeded to calculate the second-degree constant term: $$ f'(x)=\frac{1}{2+x} $$ $$ f''(x)=-(2+x)^{-2} $$ $$ f''(0)= -\frac{1}{4}$$

So I got the negative constant term and I cannot figure out what is wrong with the original statement. I must have made a mistake in my calculations, but need help in finding it.

Answer 1

The constant term is $f(0)=\ln 2>0.$

$\textbf{Update:}$

The series for $f(x)=\ln \, (2+x)$ is $$f(x)=\underbrace{\ln 2}_\textrm{const. term} + \frac{1}{2} x -\frac{1}{4}x^2 + \cdots.$$

The constant term is the first term and it is positive.

Answer 2

Your result is correct. The logarithm function strictly increases, but always slower and slower as $x$ grows. The derivative is always positive but less and less, meaning the second order term of the Taylor expansion is always negative.