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For every $k\in\mathbb{N}$, let $ x_k=\sum_{n=1}^{\infty}\frac{1}{n^2}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}. $ Calculate the limit $\displaystyle\lim_{k\rightarrow\infty}x_k$.

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    I mean maximum, there!2012-04-24

2 Answers 2

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Solution 1: By the dominated convergence theorem, we may switch the order of the limit and the sum, so we see that $\lim_{k\rightarrow\infty}x_k =0.$

Solution 2: Notice the terms are always bounded above by $\frac{1}{n^2}$. Let $\epsilon>0$, and choose $N$ such that \sum_{n=N}^\infty \frac{1}{n^2}<\epsilon. Then choose $k$ so large that $\left(1-\frac{1}{2N}\right)^{2k}\leq \frac{\epsilon}{N}.$ It then follows that $| x_k| \leq 2\epsilon,$ and the same inequality holds for all $j\geq k$. Since $\epsilon$ was arbitrary the proof is finished.

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Hint Since \displaystyle\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)<1\,\forall n\in\mathbb{Z}^+, it follows that $\displaystyle\lim_{k\rightarrow\infty}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}=0$.

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    @Eric Au contraire, I feel that this hint can be of help to the OP. Your answer, on the other hand, whilst correct, is in my opinion overly intricate for a question at this level (I'd guess it's taken from a 1st year undergrad math book).2012-04-24