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Prove that $\lim_{n \to \infty} \left(\frac{1}{n+1}\right)^2 + \left(\frac{1}{n+2}\right)^2 + \ldots = 0$. [duplicate]

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We want to prove that$$\lim_{n \to \infty} \left(\frac{1}{n+1}\right)^2 + \left(\frac{1}{n+2}\right)^2 + \dots = 0.$$

Let $S_n$ be the sum:$$S_n = \left(\frac{1}{n+1}\right)^2 + \left(\frac{1}{n+2}\right)^2 + \dots + \left(\frac{1}{2n}\right)^2.$$

Each term in the sum is of the form $\left(\frac{1}{n+k}\right)^2 = \frac{1}{(n+k)^2}$. We can bound each term:$$\left(\frac{1}{n+k}\right)^2 < \frac{1}{n^2}$$for all $k$ from $1$ to $n$. Hence,$$S_n < n \cdot \frac{1}{n^2} = \frac{1}{n}.$$

As $n$ approaches infinity, $\frac{1}{n}$ approaches $0$. Therefore, by the squeeze theorem, we can conclude that $\lim_{n \to \infty} S_n = 0$, which proves the given limit.


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