Consider the function $f(x) = \ln(1+x)$
In the textbook it is given that the Taylor's series of the function is valid for $|x|<1$ and $x=1$. $\ln(1+x) = x - (x^2)/2 + ( x^3)/3 +...... $ The nth term of the series is $\frac{(-1)^{n-1}}{n} x^n$
The nth term converges to $0$ as $n\to\infty$ if $x$ belongs to $[-0.5,1]$
But for $x$ belongs to $(-1,-0.5)$ the nth term doesn't converge to $0$, which contradicts the statement of the text book. Can someone say me where am I wrong??
In my experience, you get the most bang for your buck with these problems by doing the Ratio Test. This is a generalization of the geometric series $\sum_{n=0}^{\infty} r^n$, which converges if and only if $|r|<1$. (We'll actually make use of its closed-form later).
We must compute $$ \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty} \left|\frac{\frac{(-1)^n x^{n+1}}{n+1}}{\frac{(-1)^{n-1} x^{n}}{n}}\right| $$ $$ =\lim_{n\to\infty} \left|-x \cdot \frac{n}{n+1}\right|=|x| $$This means as long as $|x|<1$- note this implies the region $(-1,-0.5)$ that you mentioned- the series will converge. Convergence at the boundary is delicate and must be checked case-by-case. At $x=1$, we have $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}, $$which converges by the Alternating Series Test. At $x=-1$, we have $$ \sum_{n=1}^{\infty}\frac{1}{n}, $$which is the divergent harmonic series. So the series converges for $x\in (-1,1]$.
Lastly, here's a quick sketch why the series equals $\ln(1+x)$ in this region. For $|r|<1$, the geometric series yields $$ \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} $$If we put $r=-x$ we have $$ \sum_{n=0}^{\infty} (-1)^n x^n = \frac{1}{1+x} $$And now if we integrate both sides with respect to $x$ (legal for $x\in (-1,1)$) we have $$ \int\sum_{n=0}^{\infty} (-1)^n x^n \,dx = \ln(1+x) +C $$ $$ \sum_{n=0}^{\infty} (-1)^n \int x^n \,dx = \ln(1+x) +C $$ $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}x^{n+1}= \ln(1+x) +C $$A quick reindex and evaluating at $x=0$ to deduce $C=0$ gives the result: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^{n}= \ln(1+x) $$