Question in Ch-12 Apostol's Number theory (Vol1)

Question

I am trying some exercises from Apostol's Introduction to Analytic number theory and I could not solve this particular problem (number 16) of textbook and need help.

I am sorry, I wouldn't be able to provide anything as attempt as I have no ideas on which result to use.

For background, I am taking number theory course this sem where bernaulli polynomials were taught and also read them from Apostol.

Kindly shed some light on this!!

Answer 1

We start recalling the Cauchy-product when multiplying two even exponential generating functions. Given \begin{align*} A(z)=\sum_{k=0}^\infty a_{2k}\frac{z^{2k}}{(2k)!}\qquad \text{and}\qquad B(z)=\sum_{l=0}^\infty b_{2l}\frac{z^{2l}}{(2l)!} \end{align*} we obtain \begin{align*} A(z)B(z) &=\left(\sum_{k=0}^\infty a_{2k}\frac{z^{2k}}{(2k)!}\right) \left(\sum_{l=0}^\infty b_{2l}\frac{z^{2l}}{(2l)!}\right)\\ &=\sum_{n=0}^{\infty} \left(\sum_{{2k+2l=2n}\atop{k,l\geq 0}}\frac{a_{2k}}{(2k)!}\frac{b_{2l}}{(2l)!}\right)z^{2n}\tag{1.1}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{a_{2k}}{(2k)!}\frac{b_{2(n-k)}}{(2(n-k))!}\right)z^{2n}\tag{1.2}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}a_{2k}b_{2n-2k}\right)\frac{z^{2n}}{(2n)!}\tag{1.3} \end{align*} Comment:

• In (1.1) we use the Cauchy-product and sort according to even powers of $z$.

• In (1.2) we replace $l$ by $l=n-k$, observing the index range of $k$ is $0\leq k\leq n$.

• In (1.3) we use the binomial coefficient $\binom{p}{q}=\frac{(p+q)!}{p!q!}$. We see the product $A(z)B(z)$ is also an even exponential generating function.

Interpreting left-hand side and right-hand side of OPs relation as coefficients of even exponential generating functions,

we obtain according to (1.1) - (1.3) \begin{align*} \sum_{r=0}^\infty&\left(\color{blue}{\sum_{k=0}^r\frac{2^{2k}B_{2k}}{(2k)!(2r-2k+1)!}}\right)z^{2r}\\ &=\sum_{r=0}^\infty\left(\sum_{k=0}^r\binom{2r}{2k}\underbrace{2^{2k}B_{2k}}_{a_{2k}} \underbrace{\frac{1}{2r-2k+1}}_{b_{2r-2k}}\right)\frac{z^{2r}}{(2r)!}\\ &=\sum_{k=0}^\infty 2^{2k}B_{2k}\frac{z^{2k}}{(2k)!}\sum_{l=0}^\infty\frac{1}{2l+1}\,\frac{z^{2l}}{(2l)!}\\ &=\sum_{k=0}^\infty B_{2k}\frac{(2z)^{2k}}{(2k)!}\,\frac{1}{z}\sum_{l=0}^\infty\frac{z^{2l+1}}{(2l+1)!}\\ &=\left(\frac{2z}{e^{2z}-1}+z\right)\frac{1}{z}\,\sinh z\tag{2}\\ &=\left(\frac{2z}{e^{2z}-1}+z\right)\frac{1}{z}\left(\frac{e^{z}-e^{-z}}{2}\right)\\ &=\frac{e^z+e^{-z}}{2}=\cosh z\tag{3}\\ &\,\,=\sum_{r=0}^\infty\color{blue}{\frac{1}{(2r)!}}z^{2r} \end{align*} and the claim follows.

Comment:

• In (2) we recall the generating function of the Bernoulli numbers \begin{align*} \frac{z}{e^z-1}&=\sum_{r=0}^\infty B_r\frac{z^r}{r!}\\ &=1 - \frac{z}{2} + \frac{z^2}{12} - \frac{z^4}{720} + \frac{z^6}{30240}-\cdots \end{align*} where besides the linear term $-\frac{z}{2}$ all other terms with odd powers are zero. We therefore have \begin{align*} \frac{2z}{e^{2z}-1}&=\sum_{r=0}^\infty B_{2r}\frac{z^{2r}}{(2r)!}-\frac{2z}{2}\\ \frac{2z}{e^{2z}-1}+z&=\sum_{r=0}^\infty B_{2r}\frac{z^{2r}}{(2r)!} \end{align*}

• In (3) we do some simplifications.

Answer 2

Provided that Cauchy product is defined as

$$ \sum_{n=0}^\infty a_nx^n\sum_{k=0}^\infty b_kx^k=\sum_{m=0}^\infty x^m\sum_{r=0}^ma_rb_{m-r} $$

We may apply it to the left handside of what the problem states:

$$ \begin{aligned} \sum_{r=0}^\infty z^{2r}\sum_{k=0}^r{2^{2k}B_{2k}\over(2k)![2(r-k)+1]!} &=\sum_{r=0}^\infty{(2z)^{2r}B_{2r}\over(2r)!}\sum_{n=0}^\infty{z^{2n}\over(2n+1)!} \\ &=\left[{2z\over e^{2z}-1}+z\right]\cdot{e^z-e^{-z}\over2z} \\ &={e^z-e^{-z}\over e^z(e^z-e^{-z})}+{e^z-e^{-z}\over2}=\cosh z \\ &=\sum_{k=0}^\infty{z^{2r}\over(2r)!} \end{aligned} $$

Comparing the coefficients of the $z^{2r}$ terms yields

$$ \sum_{k=0}^r{2^{2k}B_{2k}\over(2k)!(2r+1-2k)!}={1\over(2r)!} $$

which is exactly what the problem asks us to prove.