Asymptotic behaviour of of $\sum_{n=0}^k a_n b_{k-n}$

Question

Let $(a_n)_n$ and $(b_n)_n$ be two sequences of positive numbers such that:

-$a_n\sim a/n^{3/2}$ for some $a>0$;

-$b_n\sim b/n^{3/2}$ for some $b>0$;

Is it true that as $k\to \infty$,

$\sum_{n=0}^k a_n b_{k-n}\sim \frac{1}{k^{3/2}}(a\sum_{n=0}^{\infty}b_n+b\sum_{n=0}^{\infty}a_n) ?$

If not true, under what additional conditions the statement is true?

Answer 1

It's true that: $\ \displaystyle \sum_{k=0}^n a_k b_{n-k}\sim \frac{1}{n^{3/2}}(a\sum_{k=0}^{\infty}b_k+b\sum_{k=0}^{\infty}a_k) $

Here is a demonstration.

$\displaystyle \sum_{k=0}^na_kb_{n-k} =t_n+u_n+v_n+w_n \quad $ with:

$t_n=\displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor}a_kb_{n-k} \quad , \quad u_n= \displaystyle \sum_{k=\lfloor \root{3}\of{n}\rfloor+1}^{\lfloor n/2\rfloor }a_kb_{n-k} \quad ,\quad v_n = \sum_{k=\lfloor n/2\rfloor +1}^{n-\lfloor \root{3}\of{n}\rfloor} a_kb_{n-k} \quad ,\quad w_n=\sum_{k=n-\lfloor \root{3}\of{n}\rfloor +1}^n a_kb_{n-k}$

• We know that there exists B such that $\ \forall n\geqslant 1 \ , \ b_n\leqslant \dfrac{B}{n^{3/2}} $. Then $ \ \displaystyle u_n\leqslant \sum_{k= \lfloor \root{3}\of{n}\rfloor+1}^{\lfloor n/2\rfloor } a_k \dfrac{B}{(n-\lfloor n/2\rfloor)^{3/2}} $. $\displaystyle 0\leqslant u_n\leqslant 2^{3/2}B \sum_{k=\lfloor \root{3}\of{n}\rfloor+1}^{+\infty}a_k \dfrac{1}{n^{3/2}} \quad $. So $\ u_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$
• By the same idea $\ v_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

$t_n-\displaystyle \dfrac{b}{n^{3/2}}\sum_{k=0}^{+\infty}a_k = x_n+y_n+z_n\quad $ with

$x_n=\displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_kb \left(\dfrac{1}{(n-k)^{3/2}}-\dfrac{1}{n^{3/2}} \right) , \quad y_n= \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_k\left(b_{n-k}- \dfrac{b}{(n-k)^{3/2}}\right), \quad z_n=-\dfrac{b}{n^{3/2}}\sum_{k=\lfloor \root{3}\of{n} \rfloor +1}^{+\infty} a_k $

• It's clear that $\ z_n={\rm o}\left(\dfrac{1}{n^{3/2}}\right)$
• $\displaystyle 0\leqslant x_n\leqslant \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_kb \left(\dfrac{1}{(n-\lfloor \root{3}\of{n}\rfloor)^{3/2}}-\dfrac{1}{n^{3/2}} \right) \leqslant \dfrac{1}{n^{3/2}} \sum_{k=0}^{+\infty}a_kb \left( \dfrac{1}{\left( 1- \frac{\lfloor \root{3}\of{n}\rfloor }{n}\right)^{3/2}}-1\right)\ $. So $\ x_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$
• $|y_n| \leqslant \displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} \dfrac{a_k}{(n-k)^{3/2}} \left| (n-k)^{3/2}b_{n-k}-b\right| \leqslant \dfrac{1}{(n-\root{3}\of{n})^{3/2}}\sum_{k=0}^{+\infty}a_k \sup_{p\geqslant n-\lfloor \root{3}\of{n}\rfloor} |p^{3/2}b_p-b|\ $ . And $\ y_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

So, we have $\ t_n \sim \dfrac{1}{n^{3/2}} b\displaystyle\sum_{k=0}^{+\infty} a_k$

With the same ideas $\ w_n \sim \dfrac{1}{n^{3/2}} a\displaystyle\sum_{k=0}^{+\infty} b_k$

And we can conclude (we may add the equivalents because they are positive).