Suppose we want to solve $$\lim\limits_{n\to \infty} \left(\left(n+1\right)^{\frac1 4}-n^{\frac1 4}\right)$$ By using series expansion. We have
$$a_n=n^{\frac1 4} \left(\left(1+\frac 1n\right)^{\frac1 4}-1\right)$$ Now let $$f:\mathbb N\setminus {1}\rightarrow (0,0.6) ~~~;n\mapsto \frac 1 n$$ and $$g(x)=(1+x)^{\frac1 4}=\sum\limits_{k=0}^{\infty} \binom {\frac 1 4}{k} ~x^k ~~;~ x\in(0,0.6)$$ Since this power series converges uniformly in $(0,0.6)$. we have composition $g\circ f$ uniformly converges on $\mathbb N\setminus {1}.$ So by uniform convergence we have $$\lim\limits_{n\to \infty} \sum\limits_{k=0}^{\infty} \binom {\frac 1 4}{k} \cdot\left(\frac {1} {n^k}\right)=\sum\limits_{k=0}^{\infty} \lim\limits_{n\to \infty} \left(\binom {\frac 1 4}{k} \cdot\left(\frac {1} {n^k}\right)\right)=1. $$ I want to use this method for
$$n^{\frac1 4} \left(\left(1+\frac 14 \cdot \frac 1n+\cdots\right)-1\right)$$. And In some book they factor out $\frac 1 n$ from series. And then applying limit term by term.
(1):My question Is Author really check that after factor out $\frac 1 n$ new series of function is uniformly convergent.?
If Answer of the question $(1)$ is yes. Then how can we justify uniform convergence quickly in such a case.
Or there is something different general theorem they are using there for applying limit term by term. That we can use in such a case.
The uniform convergence of $g\circ f$ is not sufficient, since you multiply $g\circ f(n) $by $n^{1/4}$.
When $x \in ]0,1]$, the series giving $g(x) = (1+x)^{1/4}$ is alternated and the absolute value of the general term decreases (to see that, look at the ratio of two consecutive terms). Hence the absolute value of the remainder $$g(x)-1 = \sum_{k=1}^{\infty} \binom{1/4}{k}x^k$$ is bounded above by the absolute value of $\binom{1/4}{k}x^k$, namely $x/4$. Thus, for every $n \in \mathbb{N}^*$, $n^{1/4}|g(1/n)-1| \le n^{-3/4}/4$. This gives the convergence to $0$.