# Limit of a sequence in a series form

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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Isn't it just 0? The term $(1 - \frac{1}{2n} + \frac{1}{4n^2})$ is less than $1$ for all $n$; pick any $\lambda$ between the minimum value this attains and 1; now your sum is positive but less than $\lambda^{2k}$ times the sum of $1/n^2$; since this sum converges, the limit as $k$ goes to $\infty$ is 0. –  user29743 Apr 24 '12 at 22:19
I mean maximum, there! –  user29743 Apr 24 '12 at 22:27

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 $\displaystyle\lim_{k\rightarrow\infty}\left(1-\frac{1}{2n}+\frac{1}{4n^2}\right)^{2k}=0$.
you mean $\lim_{k\rightarrow\infty}x_k=0$ –  Alex R. Apr 24 '12 at 22:22