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
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
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.
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.
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.
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 $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$
Can you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.
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
$$R_N < \frac{1}{(2n+3)!} \left( 1 + \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$
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.
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 $\sinh(1)$