hornejada1c

Answered

2022-07-11

For every $k\in \mathbb{N}$, let

${x}_{k}=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}{(1-\frac{1}{2n}+\frac{1}{4{n}^{2}})}^{2k}.$

${x}_{k}=\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}{(1-\frac{1}{2n}+\frac{1}{4{n}^{2}})}^{2k}.$

Answer & Explanation

escampetaq5

Expert

2022-07-12Added 12 answers

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 $\u03f5>0$, and choose N such that

$\sum _{n=N}^{\mathrm{\infty}}\frac{1}{{n}^{2}}<\u03f5.$

Then choose k so large that

${(1-\frac{1}{2N})}^{2k}\le \frac{\u03f5}{N}.$

It then follows that $|{x}_{k}|\le 2\u03f5,$ and the same inequality holds for all $j\ge k$. Since $\u03f5$ was arbitrary the proof is finished.

$\underset{k\to \mathrm{\infty}}{lim}{x}_{k}=0.$

Solution 2: Notice the terms are always bounded above by $\frac{1}{{n}^{2}}$. Let $\u03f5>0$, and choose N such that

$\sum _{n=N}^{\mathrm{\infty}}\frac{1}{{n}^{2}}<\u03f5.$

Then choose k so large that

${(1-\frac{1}{2N})}^{2k}\le \frac{\u03f5}{N}.$

It then follows that $|{x}_{k}|\le 2\u03f5,$ and the same inequality holds for all $j\ge k$. Since $\u03f5$ was arbitrary the proof is finished.

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