As we have the binomial theorem, $ (a+b)^n=a^n+\binom{n}{1}a^{n}b^1+\binom{n}{2}a^{n-2}b^2+\binom{n}{3}a^{n-3}b^3+..... $
This can be modified as, $ (a+b)^n=a^n(1+\binom{n}{1}(b/a)^1+\binom{n}{2}(b/a)^2+\binom{n}{3}(b/a)^3+.....) $ ......(1)
We observe that, a general Binomial coefficient term $ \binom{n}{k} $ is always there and it is independent of our choice of assigning values to '$a$' or '$b$' without considering the relation $ \mid a \mid \geq \mid b \mid $ or $ \mid a \mid \leq \mid b \mid $ . Also, it can be proved easily that $ \lim_{{k}\to{\infty}} \binom{n}{k}=0 $ for any finite 'n' .
also, from Eq.(1) , it can be easily concluded that, the sequence $ (b/a)^1, (b/a)^2, (b/a)^3 , (b/a)^4,.... $ can be convergent, only iff $ \mid b/a\mid \leq 1 $ . When $a=b$ , then this sequence converges to 1 , otherwise it converges to zero.
Hence from above discussion, it follows that whole Binomial series is a convergent sum only iff, $ \mid b/a\mid \leq 1 $ , i.e. $ \mid a \mid \geq \mid b \mid $ .
Hence, We can easily conclude that, when ' $ n $' is negative and/or fractional number, then the necessary condition for absolute convergence of $ (a+b)^n $ is $ \mid a \mid \geq \mid b \mid $ . Otherwise $ (a+b)^n $ diverges to infinity.
When, ' $ n $' is negative and/or fractional number , then the expansion of binomial theorem always gives infinite terms because there are infinite number of possibilities of $a^pb^q$ such that p+q=n and $ q $ is always a positive number.
If ' $ n $' is a positive integer, then the expansion of binomial theorem always gives finite terms because there are only finite possibilities of $a^pb^q$ such that $ p+q=n $ and hence there is no such necessary condition for absolute convergence of $ (a+b)^n $ . But still, if we wan to truncate the Binomial expansion of $ (a+b)^n $ after few terms, then in that scenario, the condition $ \mid a \mid \geq \mid b \mid $ is again must to get more and more closer to actual value of $ (a+b)^n $ . One of the best example for this is, if you expand $ \lim_{{x}\to{0}}(1+x)^\frac{1}{x} $ by using Binomial theorem, then you must expand it by observing that $ x<1 $ and hence $ a=1 $ and $ b=x $ . Then only you will get $ \lim_{{x}\to{0}}(1+x)^\frac{1}{x}=1+1+1/2!+1/3!+1/4!+.... $ . Otherwise you get terms containing the undefined quantity $ 1/0 $ .