How to show $\lim_{n\to\infty}n\cdot \sum_{m=1}^{\infty}\Big(1-\frac{1}{m}\Big)^n\cdot \frac{1}{m^2}=1.$

I am wondering if the following limit is correct and how to show it.

$$\lim_{n\to\infty}n\cdot \sum_{m=1}^{\infty}\Big(1-\frac{1}{m}\Big)^n\cdot \frac{1}{m^2}=1.$$

Edit: I apologize for the lacking context. My initial motivation was considering the probability that given $n+1$ randomly chosen positive integers, a prime $p$ divides at least two of them. I understand that “randomly choosing $n+1$integers” isn’t rigorous, and this is simply introducing how I got to the infinite series above. Assuming that a prime $p$ divides an integer with probability $1/p$ and the process is independent, the probability $f(n,p)$ is

$$f(n,p)=1–\Big(1-\frac{1}{p}\Big)^n\Big(1+\frac{n}{p}\Big).$$

Interestingly, summing $f(n,p)$ over all primes and observing the plot on mathematica, it seems to bound the prime-counting function $\pi(n)$ from below and I conjectured $\sum_{p}f(n,p) \sim \pi(n).$

Naturally, I also considered $f(n+1,p)-f(n,p),$ which is $$n\Big(1-\frac{1}{p}\Big)^n\cdot \frac{1}{p^2}.$$

Noting that the logarithmic integral approximates the prime-counting function and again comparing plots on mathematica, I conjectured the above formula summed over all primes asymptotically equals $1/\ln(n).$

Finally, to consider a simpler version, I summed the above formula over all positive integers $m$, which is the infinite series in my original question. Again based on mathematica and “feeling” based off the prime number theorem, I guessed the limit to be 1 as $n$ grows to infinity.

Solution 1:

Here is a rough idea, but by no means rigorous. Think that the sum approximates a Riemann integral with step size $\frac1n$. Then it approximates $$ \int_0^\infty \left(1-\frac1{nx}\right)^n \frac{dx}{x^2} .$$ Then the limit as $n \to \infty$ is $$ \int_0^\infty \exp\left(-\frac1x\right) \frac{dx}{x^2} = 1 .$$ But since we are working with two limiting processes at the same time, I think this answer is not rigorous.

Let's try to make it rigorous. Since $1-x \le e^{-x}$, we know that $$ n \sum_{m=1}^\infty \left(1-\frac1m\right)^n \frac1{m^2} \le n \sum_{m=1}^\infty \exp\left(-\frac nm\right) \frac1{m^2} $$ And we can definitely apply the Riemann sum idea to the right hand side. So the lim sup is bounded above by $1$.

So what about a lower bound? It seems to me (by considering Taylor's series of order 2) that there is a constant $c>0$ such that for $\alpha>0$ sufficiently small, that $(1-x) \ge e^{-(1+\alpha)x}$ for $x \in [0,c\alpha]$.

So $$ n \sum_{m=1}^\infty \left(1-\frac1m\right)^n \frac1{m^2} \ge n \sum_{m=1/(c \alpha)}^\infty \exp\left(-(1+\alpha)\frac nm\right) \frac1{m^2} . $$ The limit of this is bounded below by $$ \int_{0}^\infty \exp\left(-\frac{1+\alpha}{x}\right) \frac{dx}{x^2} .$$ Thus the lim inf is bounded below by this quantity for all $\alpha>0$, and hence the lim inf is bounded below by $1$.

Here is another approach, inspired by the answer given by @Claude Leibovici. Consider the function $$ f(x) = \frac n{n+1}\left(1-\frac1 x\right)^{n+1} .$$ Use Taylor's series of order 1 with the remainder term: $$ f(m+1) - f(m) = n\left(1-\frac1m\right)^n \frac1{m^2} + f''(\zeta_m) ,$$ where $\zeta_m \in [m,m+1]$. Sum it up, and use a telescoping series. Then you just have to estimate $$ \sum_{m=1}^\infty f''(\zeta_m) .$$ If you split the sum into two parts: $$ \sum_{1 \le m \le n^\theta} + \sum_{n^\theta < m < \infty} $$ where $\theta < 1$ is close to $1$ (probably $\theta = 9/10$ will work), you should see that this converges to $0$ as $n \to \infty$.

Solution 2:

Starting from @Stephen Montgomery-Smith's answer

$$\int\left(1-\frac1{nx}\right)^n \frac{dx}{x^2}=\frac {nx-1}{(n+1)x}\left(1-\frac{1}{n x}\right)^n$$ $$\int_0^\infty\left(1-\frac1{nx}\right)^n \frac{dx}{x^2}=\frac {n}{n+1}$$

On the other side $$\int \exp\left(-\frac{1+\alpha}{x}\right) \frac{dx}{x^2}=\frac1{1+\alpha}\exp\left(-\frac{1+\alpha}{x}\right)$$

$$\int_{0}^\infty \exp\left(-\frac{1+\alpha}{x}\right) \frac{dx}{x^2}=\frac1{1+\alpha}$$