I was doing a problem on uniform convergence. The book wishes to prove $$\frac{x^{2}\sin x}{(1+nx^{2})(1+(n-1)x^{2})}$$ uniformly converges to $x$. The first inequality is derived from the book's use of the definition of uniform convergence, which is to set $S(x) - S_{n}(x) < \epsilon$. In doing so, my book just skipped from
$$x-\frac{x^{2}\sin x}{(1+nx^{2})(1+(n-1)x^{2})}$$ to this $$\frac{|\sin x|}{1+nx^{2}}$$
I have no idea how that came about. Does anyone have any idea? I've tried basic variable manipulation, but I couldn't get anything substantial out of that denominator.
(since I don't know how this is done, I put a couple tags that may turn out to be extraneous)
To find the value of the series $$ \sum_{n=1}^\infty \frac{x^2 \sin x}{(1+nx^2)(1+(n-1)x^2)}$$ it will help to find an expression for the partial sums, and then find what the limit of the partial sums are. (And the issue of uniform convergence is then asking if the partial sums converge uniformly)
On way of doing this is to use partial fractions to decompose $$ \frac{x^2}{(1+nx^2)(1+(n-1)x^2)}$$ into partial fractions. This often allows one to write the sum as a telescoping series, which is easy to evaluate.
In this case, one finds that $$\frac{x^2 \sin x}{(1+nx^2)(1+(n-1)x^2)} = \frac{1}{1+(n-1)x^2}-\frac{1}{1+nx^2}$$
The original sum is then $$\sum_{n=1}^\infty \left(\frac{\sin x}{1+(n-1)x^2}-\frac{\sin x}{1+nx^2}\right)$$
and the partial sums are $$ S_n(x) = \sum_{k=1}^n \left(\frac{\sin x}{1+(k-1)x^2}-\frac{\sin x}{1+kx^2}\right)$$
Most of the terms cancel, and we are left with just $$ S_n(x) = \sin x - \frac{\sin x}{1+nx^2} $$
For a fixed $x$, it is not too difficult to see that $$ S_n(x) \to \sin x \text{ as } n \to \infty $$ so the pointwise limit of the series is $S(x) = \sin x$
To investigate the uniform convergence, we consider the quantity $$ |S(x) - S_n(x)| $$ which is equal to $$ \left|\sin x - \sin x + \frac{\sin x}{1+nx^2} \right| = \frac{|\sin x|}{1+nx^2} $$
The original series will converge uniformly to $\sin x$ provided that this quantity converges uniformly to $0$.
I don't know where the quantity $$x - \frac{x^2 \sin x}{(1+nx^2)(1+(n-1)x^2)}$$ comes from though. It might be because the book erroneously starts off by asserting that the series converges to $x$ instead of $\sin x$. It should instead be considering the quantity $$\sin x - \sum_{k=1}^n \frac{x^2 \sin x}{(1+nx^2)(1+(n-1)x^2)}$$ which is equal to $$ \frac{\sin x}{1+nx^2} $$ as shown above.