real analysis - Why does $sumlimits_{k=n+1}^inftyfrac{r^{2k+1}}{(2k+1)!}$ 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?

Answer

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.

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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.