Sign of remainder in series for $\cos$ and $\sin$ For real $y$, show that every remainder in the series for $\cos y$ and $\sin y$ has the same sign as the leading term.
The series for $\cos x$ is $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots$. So what we're asked to prove is that the series $\sum_{i=n}^\infty(-1)^{i}\dfrac{x^{2i}}{(2i)!}$ has the same sign as $(-1)^{n}\dfrac{x^{2n}}{(2n)!}$. Which is equivalent to $$\sum_{i=0}^\infty(-1)^i\frac{x^{2n+2i}}{(2n+2i)!}>0$$ How can one prove this?
 A: By the parity of $\cos$ and $\sin$, it is sufficient to prove it for $y > 0$. We have Taylor's formula
$$f(y) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}y^k + \frac{1}{n!} \int_0^y (y-t)^n f^{(n+1)}(t)\,dt$$
for all sufficiently smooth $f$.
For $\cos$, we choose $n = 2m$ and obtain
$$\begin{align}
\cos y &= \sum_{k=0}^m \frac{(-1)^k}{(2k)!}y^{2k} + \frac{1}{(2m)!}\int_0^y (y-t)^{2m} \cos^{(2m+1)}t\,dt\\
&= \sum_{k=0}^m \frac{(-1)^k}{(2k)!}y^{2k} + \frac{(-1)^{m+1}}{(2m)!} \int_0^y (y-t)^{2m} \sin t\,dt,
\end{align}$$
for $\sin$ we choose $n = 2m-1$ and obtain
$$\begin{align}
\sin y &= \sum_{k=0}^{m-1} \frac{(-1)^k}{(2k+1)!}y^{2k+1} + \frac{1}{(2m-1)!}\int_0^y (y-t)^{2m-1} \sin^{2m} t\,dt\\
&= \sum_{k=0}^{m-1} \frac{(-1)^k}{(2k+1)!}y^{2k+1} + \frac{(-1)^m}{(2m-1)!} \int_0^y (y-t)^{2m-1}\sin t\, dt.
\end{align}$$
So it remains to see that
$$\int_0^y (y-t)^k \sin t\, dt > 0$$
for all $y > 0$ and $k > 0$. But that follows since $(y-t)^k$ is a strictly decreasing positive function, so while $2\pi m \leqslant y$ we have
$$\int_{2(m-1)\pi}^{(2m-1)\pi} (y-t)^k\sin t\,dt > (y-(2m-1)\pi)^k\int_0^\pi\sin t\,dt > \int_{(2m-1)\pi}^{2m\pi}(y-t)^k \lvert \sin t\rvert\,dt,$$
and if $y$ is not an integer multiple of $2\pi$, the integral of the part of the last period is still positive (clear when $2m\pi < y < (2m+1)\pi$, by the analogous estimate
$$\begin{align}
\int_{2m\pi}^y (y-t)^k\sin t\,dt &= \int_{2m\pi}^{(2m+1)\pi}(y-t)^k\sin t\,dt - \int_{(2m+1)\pi}^y(y-t)^k\lvert\sin t\rvert\,dt\\
&> (y-(2m+1)\pi)^k \int_0^\pi\sin t\,dt - (y-(2m+1)\pi)^k\int_0^{y-2m\pi}\sin t\,dt\\
&> 0
\end{align}$$
when $(2m+1)\pi \leqslant y < 2(m+1)\pi$).
A: This result is valid for an alternating series $\sum (-1)^n a_n$ (or $\sum (-1)^{n+1} a_n$) where $(a_n)$ is a decreasing sequence to $0$.
Let
$$s_n=\sum_{k=0}^n (-1)^k\frac{x^{2k}}{(2k)!}=\sum_{k=0}^n u_k$$
and 
$$S=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$$
so we can see that the sequence $(s_{2n})$ is decreasing and the series $(s_{2n+1})$ is increasing and are convergent to $S$ so we have
$$s_{2n+1}\leq S\leq s_{2n+2}\leq s_{2n}$$
hence 
$$u_{2n+1}\leq r_{2n+1}=S-s_{2n}\leq 0\tag{1}$$
and
$$0\leq r_{2n+2}=S-s_{2n+1}\leq u_{2n+2}\tag{2}$$
so from $(1)$ and $(2)$ we see that the remainder $r_n$ has the sign of the term $u_n$ and that
$$|r_n|\leq |u_n|$$
