Using Ratio Test for Taylor Series of Natural Log At $x = 1$ I was told to apply the Ratio Test for the Taylor series based at $x=1$ for $ln(x)$. Using the test I have to show that the series converges when $0 < x \le 1$. 
My approach:
When I did the ratio test by taking the limit as $n$ approaches infinity of $|\frac{(-1)^{n-1}\frac{(x-1)^n}{n}}{(-1)^{n-2}\frac{(x-1)^{n-1}}{n-1}}|$, the value I got was $|{x-1}|$. I know that the limit should be less than $1$, so I evaluated $|x-1| < 1$, for which I got $0 < x < 2$, showing that $x$ needs to be in that interval for the series to converge -- different from what they told me. 
 A: If $x\leq 1,$ then $x\leq 2.$
Your argument, which is sufficient to prove convergence for $0<x\leq 2,$
is certainly also sufficient to prove convergence for $0<x\leq 1.$
After all, there is no $x$ in the interval $(0,1]$ that is not in the interval $(0,2].$
It is not true that the series converges if and only if $0<x\leq 1.$
But you were not asked to prove that. You were only asked to prove the “if” part. 
And the “if” part is true. 
One does have to wonder why the exercise asked for a proof of a theorem so much weaker than the one you are able to prove. Usually we would want the stronger theorem in a case like this where the stronger statement is just as simple as the weaker one. Possibly the author had in mind that you could use a weaker argument (but I have no guess what that would be), or perhaps there will be a need later to know this for $0<x\leq1$ and this led the author not to think about the possibility of convergence on a larger interval. Or perhaps it was just a transcription error and it should have been $2$ instead of $1$. But I am just guessing now. 

Another way to think about it is, suppose you were asked to show that the series converges when $x=\frac12$?
Surely you could do that. What the question actually asked is somewhere in between asking you to show convergence for just one value, and asking you to find all values of $x$ for which the series converges. 
