Question

Why for some McLaurin expansions like \(\large\rm (1+x)^{-1}, ln(1+x)\ \) the value of x cannot be greater than 1?

Answer 1

Because they diverge

Answer 2

Look up "radius of convergence", or
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

Answer 3

they are only valid when the series converges

Answer 4

\(\color{black}{\displaystyle (1-x)^{-1}=\sum_{n=1}^\infty x^n}\)

\(\color{black}{\displaystyle (1+x)^{-1}=\sum_{n=1}^\infty (-x)^n}\)

If \(\color{black}{\displaystyle \left|x\right|\ge1}\), then the series diverges. (An elementary geometric series ratio requirement for convergence is not satisfied under the aforementioned condition.)

\(\color{black}{\displaystyle \ln(1+x)=\int (1+x)^{-1}dx=\int \sum_{n=1}^\infty (-1)^nx^n~dx=x\cdot \sum_{n=1}^\infty (-1)^nnx^{n}}\)

and then again you need \(\color{black}{\displaystyle \left|x\right|<1}\) for convergence.

Answer 5

The question about ln(x+1), however is a bit harder. When |x|<1, you are having a(n increasing) exponential function in the denominator, and n in the numerator.

Answer 6

You can actually prove that any
\(\displaystyle \sum_{n=j}^{\infty}\frac{\rm increasing~power~function}{\rm exponential~growth~function}\)
will always diverge, or just abide by this rule as a common sense.