2022-07-11

For every $k\in \mathbb{N}$, let
${x}_{k}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}{\left(1-\frac{1}{2n}+\frac{1}{4{n}^{2}}\right)}^{2k}.$

escampetaq5

Expert

Solution 1: By the dominated convergence theorem, we may switch the order of the limit and the sum, so we see that
$\underset{k\to \mathrm{\infty }}{lim}{x}_{k}=0.$
Solution 2: Notice the terms are always bounded above by $\frac{1}{{n}^{2}}$. Let $ϵ>0$, and choose N such that
$\sum _{n=N}^{\mathrm{\infty }}\frac{1}{{n}^{2}}<ϵ.$
Then choose k so large that
${\left(1-\frac{1}{2N}\right)}^{2k}\le \frac{ϵ}{N}.$
It then follows that $|{x}_{k}|\le 2ϵ,$ and the same inequality holds for all $j\ge k$. Since $ϵ$ was arbitrary the proof is finished.

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