$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges? Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Thought about using the limit theorem or by comparison but am so stuck.  any pointers would be appreciated guys
 A: I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is  obviously a number, so he doesn't have to prove that the series for e converges.
Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

Therefore, the series converges.
A: The series you have is 
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots $
If you add the even factorial terms, you get an upper bound i.e.,
$1 + \frac{1}{3!} + \frac{1}{5!} \ldots < \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$
This can be written more compactly as
$\sum_{n=0}^\infty \frac{1}{(2n+1)!} < \sum_{n=0}^\infty \frac{1}{n!} = e^1$
Therefore the series converges.
A: We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots $$
Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
A: Can you bound the series from above by one that you know converges?  The factorials grow very fast, so you should be able to.
A: To elaborate on the first answer given to this question by Ross Millikan.
$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$
$$< 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } 
\frac{1}{(2n+1)!} < \frac{1}{4^n} \textrm{ for } n \ge1.$$
Hence by the comparison test the series converges.
Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that
$$R_N < \frac{1}{(2n+3)!} 
\left( 1 +  \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$
A: Another way is
If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$
We have that
$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$ 
$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$
Thus $S_n < 2$ 
thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.
A: METHOD I
We may simply resort to the Basel problem and get the inequality:
$$0<\sum_{k=0}^{\infty}\frac{1}{(1+2k)!}\leq\sum_{k=0}^{\infty}\frac{1}{(1+k)^2}=\frac{\pi^2}{6}$$
METHOD II
According to Taylor's expansion we have that:
$$ \sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!}$$
For $x=1$ we get that the value of the series is $\sinh(1)$. The series converges.
Q.E.D. 
A: This is how sometimes we can extract the exact closed form of some series : 
$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$
The radius of convergence is $\infty$
$f \in C^{\infty}(\mathbb{R})$
$f''(x) = f(x)$
solving our equation, we get $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$
$\lambda^2-1=0$ 
$\lambda \in \{-1,1\}$
$f(x) = ae^{-x}+be^{x}$
$f(0)= 0 \implies a+b=0 \implies a=-b \implies f(x) = a(e^{x}- e^{-x})$
$f'(0) = 1 \implies a=\frac{1}{2}$
$f(x) = \frac{e^{x}-e^{-x}}{2}$
so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$
