Decay rate of tensor product of series Suppose I have a series $a_n = n^{-1}$. Now I define a double-indexed series $b_{m,n}:=a_m\cdot a_n$. And I reindex $b_{m,n}$ to a signle-indexed series $c_n$ so that $c_n$ is decreasing w.r.t. $n$. For example, the first several elements of $c_n$ is
\begin{equation}
c_1 = 1;\ c_2=c_3=1/2;
\end{equation}
Can I tell how $c_n$ depends on $n$ asymptotically? Is it $c_n\sim n^{-1}$? $c_n\sim n^{-2}$? or $c_n\sim exp(-n)$?
 A: For convenience of notation I am going to reindex the sequence so that $c_0 = 1, c_1 = c_2 = 1/2$, etc.
First notice that the sequence $(c_n)$ consists only of numbers of the form $1/m$, $m \in \mathbb{N}$, and that $1/m$ appears in the sequence exactly $d(m)$ times, where $d(m)$ is the number of divisors of $m$.  Therefore we can describe the general term of the sequence $(c_n)$ as
$$
c_n = \frac{1}{k(n)},
$$
where $k(n)$ is defined to be the largest $k$ such that $\sum_{m=1}^{k} d(m) \leq n$.
Now we can use the well-known elementary estimate $\sum_{m=1}^{k} d(m) = k\log k + O(k)$ (see for example here) to establish the asymptotic $k(n) \sim \frac{n}{\log n}$.  First we note that
$$
\sum_{m=1}^{n/\log n} d(m) = \frac{n}{\log n} \log \left( \frac{n}{\log n} \right) + O\left(\frac{n}{\log n}\right) = n - n \frac{\log\log n}{\log n} + O\left(\frac{n}{\log n}\right) < n
$$
for $n$ sufficiently large, thus $k(n) \geq \frac{n}{\log n}$.  For the reverse inequality, suppose that $k(n) > (1+\epsilon) \frac{n}{\log n}$ for infinitely many $n$.  Then we would have
\begin{align*}
n &\geq \sum_{m=1}^{(1+\epsilon)n/\log n} d(m) = (1+\epsilon) \frac{n}{\log n} \log \left( (1+\epsilon) \frac{n}{\log n} \right) + O\left(\frac{n}{\log n}\right) \\
&=  (1+\epsilon)n - (1+\epsilon) n \frac{\log \log n}{\log n} + O\left(\frac{n}{\log n}\right) \\
&= (1+\epsilon)n + o(n),
\end{align*}
a contradiction once $n$ is sufficiently large.
Thus we have established $k(n) \sim \frac{n}{\log n}$ and therefore the desired asymptotic is
$$
c_n \sim \frac{\log n}{n}.
$$
