Let $\lambda\vdash n$ denote the integer partition of $n$. Define the product $\mathcal{N}(\lambda)=\lambda\_1\lambda\_2\cdots\lambda\_r$ when $\lambda=(\lambda\_1\geq\lambda\_2\geq\cdots\geq\lambda\_r>0)$.


Let $\gamma$ be the Euler's constant. Lehmer proved that
$$\lim\_{n\rightarrow\infty}\frac1n\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)}=e^{-\gamma}.$$


I like to ask:



> 
> **QUESTION.** Does the following limit exist? Or, what is the asymptotic growth of the sum?
> $$\lim\_{n\rightarrow\infty}\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}.$$
> 
> 
> 


**ADDED.** Glad to see that Fedor Petrov's answer would (potentially) work for
$$\lim\_{n\rightarrow\infty}\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^r}.$$



asymptotic growth of a sum involving partitions
The limit equals 2. We have $$
\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}=[x^n]\prod\_{k=1}^\infty\frac1{1-x^k/k^2}=\sum\_{m=0}^n[x^m]\prod\_{k=2}^\infty\frac1{1-x^k/k^2}\\
=b\_0+b\_1+\ldots+b\_n,$$
where
$$
\sum b\_i x^i=\prod\_{k=2}^\infty \frac1{1-x^k/k^2}=:f(x),
$$
the standard uniform convergence argument shows that $$\sum b\_i=\sup\_{0<x<1} f(x)=f(1)=\prod\_{k=2}^\infty\frac{k^2}{k^2-1}=2.$$


