Why does $\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}$ converge to $0$? I'm being told that because the following series is the tail end of a convergent series, it converges to zero as $n$ gets large:

$$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}$$

The tail end of which convergent series? $e^r$? If so, then the above series is actually every other term of the tail send of the power series for $e^r$, right?
Or how else to see that the above series converges to $0$? Or does the series sum to zero simply because as $n$ gets large, the number of terms get arbitrarily small?
 A: The series $\displaystyle \sum_{k=0}^\infty \frac{r^{2k+1}}{(2k+1)!}$ is the Taylor series of $\sinh r$, the hyperbolic sine (see Wikipedia), which converges for all real $r$.
But even the observation that it's half of the terms of $e^r$ is sufficient in this case: simply apply the squeeze theorem with the zero series and the $e^r$ series:
In terms of the formulation in the link, put:
\begin{align}
x_n &= \sum_{k=n+1}^\infty \frac{r^{2k+1}}{(2k+1)!}\\
y_n &= 0\\
z_n &= \sum_{k=n+1}^\infty \frac{r^{2k+1}}{(2k+1)!} + \sum_{k=n+1}^\infty \frac{r^{2k+2}}{(2k+2)!} = \sum_{j=2n+3}^\infty\frac{r^j}{j!}
\end{align}
Since $z_n$ is the tail of $e^r$, it converges to zero, and the squeeze theorem tells us that $\lim\limits_{n\to\infty} x_n = 0$, as desired.

As to "the number of terms gets arbitrarily small", that's quite incorrect. It's like saying that there are only finitely many natural numbers.
If this were correct, we would have $\displaystyle \sum_{k=n+1}^\infty 1 = 0$ as well, which of course is not true.
A: Quotient $|a_{k+1}/a_k| \leq \delta < 1$ for $k$ sufficiently large wrt to $r$.
A: Let $x_n = \sum_{k=n+1}^\infty a_k$ (if it exists).
if $x_0 = \sum_{k=1}^\infty a_k$ exists, by definition this means that the sequence $b_n = \sum_{k=1}^n a_k$ has a limit $l$. Since $x_n = l - b_n$ and $b_n \to l$, $x_n$ must converge to $0$.
So the sequence $(x_n)$ converges to $0$ because it's in the definition of "$x_n$ exists".
So I guess if you're even talking about your sequence, hopefully you know that it exists, and so the sequence converge to $0$. If not the problem is to show that $x_0$ exists, and you do that by comparing the finite sums to $e^r$ or to some other converging series like $\sum a2^{-k}$ for some big enough $a$
