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$$I=\lim_{n\to\infty}\sum_{k=0}^{n}\dfrac{\cos{\sqrt{\dfrac{k}{n}}}}{2^k}$$

my idea: since $$\cos{\sqrt{\dfrac{k}{n}}}\le 1$$ so $$I\le\lim_{n\to\infty}\sum_{k=0}^{n}\dfrac{1}{2^k}=2$$

But How show $I\ge 2?$ Thank you

math110
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    isn't $I \le 2$ ?? – S L Jul 11 '14 at 07:52
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    I was going to post a comment on an answer that is deleted, but maybe it is still of use. To give a warning about why you can't just plug in $n=\infty$ everywhere, consider $\lim\limits_{n\to\infty}\sum\limits_{k=0}^n\dfrac{1}{k+n}$. This limit is not $0$. – Jonas Meyer Jul 11 '14 at 08:20

2 Answers2

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Note that $\forall x\in\mathbb{R},~~1-\frac{x^2}{2}\le\cos{x}$. Thus,

$$\cos{\sqrt{\frac{k}{n}}}\ge 1-\frac{k}{2n}\\ \implies I\ge \lim_{n\to\infty}\sum_{k=0}^{n}\frac{1-\frac{k}{2n}}{2^k}\\ =\lim_{n\to\infty}\left[2+\frac{2^{-n}-1}{n}-2^{-n-1}\right]\\ =2.~~~\blacksquare$$

David H
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Here is an alternate way to evaluate the limit uses monotone convergence theorem (for sequences).

Notice the partial sums can be rewritten as

$$\sum_{k=0}^n 2^{-k} \cos\sqrt{\frac{k}{n}} = \sum_{k=0}^\infty \alpha_{n,k}\quad\text{where}\quad \alpha_{n,k} = \begin{cases}2^{-k}\cos\sqrt{\frac{k}{n}},&n \ge k\\0, & n < k\end{cases}$$ For fixed $k$ and as a function of $n$, $\alpha_{n,k}$ is non-negative and monotonic increaing. By MCT, you can switch the order of taking limit and sum and get

$$\lim_{n\to\infty}\sum_{k=0}^n 2^{-k}\cos\sqrt{\frac{k}{n}} = \lim_{n\to\infty}\sum_{k=0}^\infty \alpha_{n,k} = \sum_{k=0}^\infty \lim_{n\to\infty}\alpha_{n,k} = \sum_{k=0}^\infty 2^{-k} =2$$

achille hui
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  • I'm curious, does what's going on in your reponse explain on some deep level how I got away with such a cheap derivation of the greatest lower bound? Because honestly, I began that computation with the aim of just establishing a lower bound, expecting that I'd have to use increasingly higher order Taylor approximations to approach the glb. – David H Jul 12 '14 at 10:06
  • @DavidH The key is the coefficients involved are non-negative and monotonic increasing. You can recast the limit into the form of a double sum of non-negative numbers $$\sum_{k=0}^\infty\sum_{n=0}^\infty \left(\alpha_{n,k}-\alpha_{n-1,k}\right)$$ Since all the terms involved have the same sign, their contribute won't interfere with each other. No matter what order you choose to evaluate the sum, you always get back the same number. – achille hui Jul 12 '14 at 10:51