<s>[INST] Show that this limit yields $\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k= n}^{(n-1)^2}\frac{1}{k}$ $\gamma =0.5772156...$ is Euler's constant
Show that this limit yields
$$\gamma=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=n}^{(n-1)^2}\frac{1}{k}$$
 [/INST] Since
$$\frac{1}{k} \leqslant \int_{k-1}^k \frac{dx}{x} = \log k - \log(k-1) \leqslant \frac{1}{k-1},$$
it follows that
$$\sum_{k = n}^{(n-1)^2}\frac{1}{k} \leqslant \log(n-1)^2 - \log(n-1) = \log(n-1) \leqslant \sum_{k = n}^{(n-1)^2}\frac{1}{k} + \frac{1}{n-1} - \frac{1}{(n-1)^2},$$
and
$$ \sum_{k = 1}^{n-1}\frac{1}{k} - \log(n-1) + \frac{1}{n-1} - \frac{1}{(n-1)^2}\leqslant \sum_{k = 1}^{n-1}\frac{1}{k} - \sum_{k = n}^{(n-1)^2}\frac{1}{k} \leqslant \sum_{k = 1}^{n-1}\frac{1}{k} - \log(n-1).$$
By the squeeze theorem
$$\lim_{n \to \infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k= n}^{(n-1)^2}\frac{1}{k}\right) = \lim_{n \to \infty}\left(\sum_{k=1}^{n-1}\frac{1}{k} - \log(n-1)\right) = \gamma$$
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