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Since $X$ is discrete, you can simplify a little: $$\lim_{n\to\infty}p(X_n=0) = \lim_{n\to\infty}\text{e}^{-{1 \over n}} = \text{e}^{\lim_{n\to\infty}{-{1\over n}}} = \text{e}^0=1$$ where we can go from the second to the third term by the continuity of the exponentiation function. The second statement follows from the first, as $n\cdot0 = 0$ and $n\cdot X ... 5 Stirling's approximation gives $$\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right)$$ so $$\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{... 5 Does this mean the warning is a "false positive" and can safely be ignored ? No. A singular fit is quite specifically defined, at least in lme4, which I assume is what you are using. The warning in lme4 comes from a principal components analysis of the variance-covariance matrix of estimated random effects. If this is not of full rank, then the fit is ... 2 I assume X_{i} are all non-negative. (There is a way to treat general X_{i}, but then we need to delve into how we define expectations and such). This can be shown using the second Borel-Catelli lemma. The lemma states that given E_{1},E_{2},\ldots independent events such that \sum_{n=1}^{\infty} P(E_{n}) = \infty, then P(\limsup_{n} E_{n}) = 1. ... 2 If we redesign your sequence as Y_1=1, and everything else remains the same, the contradiction disappears. Then, if Y_n converges, it should converge to 0. Here, we'll use the definition of convergence in probability we need to show the following:$$\lim_{n\rightarrow\infty}P(|Y_n-0|>\epsilon)=\lim_{n\rightarrow\infty} P(Y_n>\epsilon)=0$$Y_n ... 1 The same question can be asked in principle for any machine learning method. We almost never decay the step-size, and in fact we often use optimizers (like ADAM) that have errors in their proof of convergence. In fact, we continue to use ADAM instead of alternatives that correct this deficiency and provably converge (see here). Why the discrepancy? Perhaps ... 1 While this is not as elementary as Stirling's approximation, the pointwise convergence of the density can be shown using dominated convergence theorem. The density of a t-distribution with n degrees of freedom is of the form$$f_n(x)=c_n\cdot\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}\quad,\,x\in\mathbb R$$Let$g_n(x)=\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}$... 1 A generalization uncovers a fundamental idea. One nice thing about it is how it circumvents calculation altogether: the Gamma functions don't play any role and, in fact, neither do the specific expressions for the Normal and Chi-squared pdfs. Recall that the Student$t$distribution with$\nu\$ degrees of freedom originates (both historically, pedagogically,...