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.

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

$$ b_n = [x^{2n}]\cosh^2(x) = \frac{2^{2n-1}}{(2n)!}.\tag{4}$$

  • $\begingroup$ 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!! $\endgroup$ – Old mate May 13 '16 at 14:07
  • $\begingroup$ @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 $0$. $\endgroup$ – Jack D'Aurizio May 13 '16 at 14:11
  • $\begingroup$ @Oldmate: anyway, my $(4)$ now provides an explicit computation of $a_n$ and $b_n$ that does not require the identity $\cosh^2-\sinh^2=1$. $\endgroup$ – Jack D'Aurizio May 13 '16 at 14:12
  • $\begingroup$ Thanks heaps - I didn't even consider differentiating. $\endgroup$ – Old mate May 13 '16 at 14:14

Hint: \begin{align} \sum_{k=0}^{2n+2}\frac{(-1)^k}{k!(2n+2-k)!} =\frac{1}{(2n+2)!}\sum_{k=0}^{2n+2}(-1)^k{2n+2\choose k} =\frac{1}{(2n+2)!}(1-1)^{2n+2}=0. \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.