Question: The convergence of \(\sum_{n=1}^{\infty}\left[(-1)^{n-1} x^{n}\right] / n\) to \(\ln (1+x)\) for \(-1

1. Show by long division or otherwise that \(\frac{1}{1+t}=1-t+t^{2}-t^{3}+\cdots+(-1)^{n} t^{n}+\frac{(-1)^{n+1} t^{n+1}}{1+t}\)
2. By integrating the equation of part (a) with respect to \(t\) from 0 to \(x\), show that
   \[\begin{aligned} \ln (1+x)=& x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \\ &+(-1)^{n} \frac{x^{n+1}}{n+1}+R_{n+1} \end{aligned}\]
   where
   \[R_{n+1}=(-1)^{n+1} \int_{0}^{x} \frac{t^{n+1}}{1+t} d t\]
3. If \(x \geq 0\), show that
   \[\left|R_{n+1}\right| \leq \int_{0}^{x} t^{n+1} d t=\frac{x^{n+2}}{n+2} \text {. }\]
   Hint: As \(t\) varies from 0 to \(x\),
   \[1+t \geq 1 \text { and } t^{n+1} /(1+t) \leq t^{n+1},\]
   and
   \[\left.\left|\int_{0}^{x} f(t) d t\right| \leq \int_{0}^{x}|f(t)| d t .\right)\]
4. If \(-1 \[\left|R_{n+1}\right| \leq\left|\int_{0}^{x} \frac{t^{n+1}}{1-|x|} d t\right|=\frac{|x|^{n+2}}{(n+2)(1-|x|)}\]
   Hint: If \(x \[\left.\left|\frac{t^{n+1}}{1+t}\right| \leq \frac{|t|^{n+1}}{1-|x|} \cdot\right)\]
5. Use the foregoing results to prove that the series

converges to \(\ln (1+x)\) for \(-1

converges if \(p>1\) and diverges if \(p \leq 1\).

Price: $2.99

Solution: The downloadable solution consists of 3 pages
Deliverable: Word Document