Why does Taylor series converge in a circle? Say there is a series $ y = \sum a_{n} x^{n}$.. Why does it converge for both positive and negative values ? Whats the intuition behind saying that ?
 A: An entire series converges when its general term decays sufficiently quickly. And if it decays in the positives, it decays the same way in the negatives. This is what happens inside the radius of convergence.
For example, the series
$$\sum\frac{(-1)^{n-1}x^n}{n}$$
is such that for $x=\pm0.5$, the $10^{th}$ term is $0.00009765625$, which is small, and the $20^{th}$ term is $2048$ times smaller.

On the opposite, on the radius of convergence, anything can occur.
For instance the series may converge in the positives because it has alternating signs and the term somewhat compensate each other to ensure sufficient decay, and diverge in the negatives because the alternating signs disappear and the decay is slower.
With the given series, the radius is $1$ and convergence is obtained for $x=1$: thanks to the sign alternation, every other term cancels out,
$$\frac1{2n}-\frac1{2n+1}=\frac1{2n(2n+1)}\approx\frac1{4n^2}$$ so that the decay is in fact quadratic.
$$\frac1{10}-\frac1{11}=0.00909090\cdots$$
$$\frac1{20}-\frac1{21}=0.00238095\cdots$$
But for $x=-1$ the decay is only harmonic,
$$\frac1{2n}+\frac1{2n+1}=\frac{4n+1}{2n(2n+1)}\approx\frac1n$$ 
and too slow
$$\frac1{10}+\frac1{11}=0.19090909\cdots$$
$$\frac1{20}+\frac1{21}=0.09761904\cdots$$

By definition, on the radius of convergence, you are at the border between convergent and divergent, and sign matters.
Inside the radius of convergence, fast decay is sign-independent (and also argument-independent for the complex).
A: For $\sum_na_nx^n$ to converge, we certainly must have  $|a_nx^n|\to 0,$ so convergence will fail if, for some  $k>0,$ we have  $|a_nx^n|\geq k$ for infinitely many $n.$ So for any $k>0$ we must have $|a_n|^{1/n}\cdot |x|<k^{1/n}$ for all but finitely many $n.$ 
Now $k^{1/n}\to 1$ as $n\to \infty$ for any $k>0$, so we must have $|a_n|^{1/n}\cdot |x|\leq 1$ for all but finitely many $n.$ Therefore convergence requires $|x|\leq R$ where $$\frac {1}{R}=\lim_{m\to \infty}\sup_{n>m}|a_n|^{1/n}.$$ 
This suggests that strict inequality $|x|<R$ may be sufficient for convergence, and it is.
In the case where $0<R<\infty,$ suppose $|x|<R$. Let $|x|=(1-a)R$ where $0<a\leq 1.$ The definition of $R$, above, implies that for any $b>0$ there are only finitely many $n$ such that $|a_n|^{1/n}\geq (1+b)\frac {1}{R}.$  So if we take $b>0$ where $b$ is small enough that $(1-a)(1+b)<1$ then  for all but finitely many $n$ we have $$|a_nx^n|\leq (1+b)^n R^{-n}\cdot |x^n|=(1+b)^n R^{-n}\cdot (1-a)^nR^n=((1+b)(1-a))^n.$$ Since $0\leq (1+b)(1-a)<1,$ this means that the absolute values of the terms $a_nx^n$ go to $0$ at least as fast as the terms of the convergent geometric series $\sum_n((1+b)(1-a))^n$, so $\sum_na_nx^n$ converges.
