# Mathematical Induction with series and factorials.

I wish to show the following

$$a_{n}=\sum_{k=0}^{n}\frac{1}{(2k+1)!(2(n-k)+1)!}=\sum_{k=0}^{n+1}\frac{1}{(2k)!(2(n+1-k))!}=b_{n+1}$$

for $n\geq0$ and wish to do it using induction. I've shown it to be true when $n=0$, no issues there. But I'm running into all sorts of trouble when I wish to show $a_{m+1}=b_{m+2}$, assuming $a_{m}=b_{m+1}$ for some $m\geq0$.

I won't write down my scribblings here, as it would honestly just be super messy. My question is, can this be done using induction? And if not, is there a simple* method that will allow me to prove the result?

*as in, under-grad simple. I'm not equipped with a whole lot of fancy theorems.

My main attempt I will write down, in case it is on the right track or helpful in any way.

Using my inductive hypothesis, I can write \begin{align*} \sum_{k=0}^{m}\frac{1}{(2k+1)!(2(m-k)+1)!}&=\sum_{k=0}^{m+1}\frac{1}{(2k)!(2(m+1-k))!}\\ &=\sum_{k=0}^{m}\frac{1}{(2k)!(2(m+1-k))!}+\frac{1}{(2(m+1))!}\\ \end{align*} My idea with this was originally to get as much as I could under one summation, and I ended up with $$\sum_{k=0}^{m}\left[\frac{1}{(2k+1)!(2(m-k)+1)!}-\frac{1}{(2k)!(2(m+1-k))!}\right]=\frac{1}{(2(m+1))!}$$ which became $$\sum_{k=0}^{m}\left[\frac{2(m-k)+2-(2k+1)}{(2k+1)!(2(m+1-k)!}\right]=\frac{1}{(2(m+1))!}$$ and $$\sum_{k=0}^{m}\left[\frac{2m-4k+1}{(2k+1)!(2(m+1-k)!}\right]=\frac{1}{(2(m+1))!}$$ but this seems way off-track. As I mentioned, I don't know whether this can be done using induction, but I'd really like it to be able to be. The RHS of the statement I wish to prove has that pesky $n+1$ term which also alters the summand. Because of this, I have no idea how to massage it to use my inductive hypothesis.

Here are some links to WolframAlpha also.

When $n=0$

When $n=1$

When $n=2$

When $n=3$

Any insight would be greatly appreciated. Thanks in advance for your time.

• Does $$a_n = \frac{1}{(2n+2)!} \sum_{k = 0}^{n} \binom{2n+2}{2k+1}$$ help? – Daniel Fischer May 13 '16 at 13:55
• I didn't see this! I'll try it out and see. Thanks! – Old mate May 13 '16 at 13:56
• And use the binomial expansion for $(1-1)^{2n+2} = 0$. – Daniel Fischer May 13 '16 at 14:11

Both $a_n$ and $b_{n}$ are given by convolutions: $$\begin{array}{cclcl} a_n &=& \displaystyle\sum_{a+b=n}\frac{1}{(2a+1)!}\cdot\frac{1}{(2b+1)!} &=& \displaystyle[x^n]\left(\sum_{c\geq 0}\frac{x^c}{(2c+1)!}\right)^2 \\ b_n &=& \displaystyle\sum_{a+b=n}\frac{1}{(2a)!}\cdot\frac{1}{(2b)!} &=& \displaystyle[x^n]\left(\sum_{d\geq 0}\frac{x^d}{(2d)!}\right)^2\end{array}\tag{1}$$ hence: $$a_n = [x^n]\left(\frac{\sinh \sqrt{x}}{\sqrt{x}}\right)^2 = [x^n]\frac{\sinh^2(\sqrt{x})}{x} \tag{2}$$ as well as: $$b_n = [x^n]\left(\cosh(\sqrt{x})\right)^2 = [x^n]\cosh^2(\sqrt{x}) \tag{3}$$ and the claim ($a_n=b_{n+1}$) just follows from the identity $\cosh^2(z)-\sinh^2(z)=1$.
In a explicit way: $$a_n = [x^{n+1}]\sinh^2(\sqrt{x}) = [x^{2n+2}]\sinh^2(x) = \frac{2^{2n+1}}{(2n+2)!},$$
$$b_n = [x^{2n}]\cosh^2(x) = \frac{2^{2n-1}}{(2n)!}.\tag{4}$$
• Haha, believe it or not, this is what sparked my question. I need to prove using power series that $\cosh^{2}(x)-\sinh^{2}(x)=1$, and as we haven't been given Cauchy product in class, I'm unable to use that for assessment. But thank you nonetheless!! – Old mate May 13 '16 at 14:07
• @Oldmate: for such a task, it is way easier to check for first that $$\frac{d}{dx}\left(\cosh^2(x)-\sinh^2(x)\right) = 0,$$ then evaluate $\cosh^2-\sinh^2$ at $0$. – Jack D'Aurizio May 13 '16 at 14:11
• @Oldmate: anyway, my $(4)$ now provides an explicit computation of $a_n$ and $b_n$ that does not require the identity $\cosh^2-\sinh^2=1$. – Jack D'Aurizio May 13 '16 at 14:12