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The following problems appears in Makarov, B. M. , et al, Selected Problems in Real Analysis, Translation of Mathematica Monographs, AMS, 1992.

Define the sequences

  1. $x_n=\sum^{2n}_{k=0}2^{-k}\cos(\sqrt{k/n})$
  2. $y_n=\sum^{2n}_{k=0}2^{-\tfrac{nk}{n+k}}$

Determine the limits of $x_n$ and $y_n$.

Armed with the tools of integration theory (dominated convergence), it is possible to solve this problems rather easily. For example, in (1) we may consider the finite measure space $(\mathbb{Z}_+,\mathcal{P}(\mathbb{Z}_+),\mu)$ where $$\mu=\sum^\infty_{k=0}2^{-k}\delta_k$$ The sequence $f_n(x)=\cos(\sqrt{x/n})\mathbb{1}_{\{0,\ldots,2n\}}(x)$ satisfies $|f_n|\leq 1$ and $\lim_nf_n=\cos(0)$. Hence $$x_n=\sum^{2n}_{k=0}2^{-k}\cos(\sqrt{k/n})=\int f_n\,d\mu\xrightarrow{n\rightarrow\infty}2$$

Question: The set of problems containing the exercise above seemed to be for students with good Calculus knowledge (sequences, differentiation and Riemann integration). My question is whether someone present a solution, albeit no necessarily easier, using the tools of College Calculus to either of these two exercises.

Edit An elementary solution to the the limit in (2)- elementary in the sense that only basic methods from Calculus- is here.

Mittens
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  • Do you allow Riemann sums? – Stefan Lafon Aug 19 '21 at 22:25
  • @StefanLafon: It could be, anything within the typical College Calculus syllabus. I though about Riemann sums for a while but did not see a clear path. – Mittens Aug 19 '21 at 22:27
  • College calculus student here: how is it that $\lim_{n\to\infty}f_n=\cos(1)$, when $\lim_{n\to\infty}\cos(\sqrt{x/n})=\cos(0)=1$ since $\sqrt{x}$ and $\cos$ are continuous, the limit can be passed all the way inside... – FShrike Aug 19 '21 at 22:28
  • From my very short education on Lebesgue theory, I gather that your indicator function (I assume that’s what $\Bbb{1}$ is) is an indicator for the integers ${0,1,\ldots,2n}$ which means $f_n$ is only non-zero for integer $x$ - is this correct? – FShrike Aug 19 '21 at 22:31
  • @FShrike: correct. I prefer the probabilist notation for the indicator function of a set. – Mittens Aug 19 '21 at 22:34
  • @JeanL. You can consider the counting measure $\nu=\sum^\infty_{k=0}\delta_k$. $g_n(x)=2^{-\tfrac{x}{1+\tfrac{x}{n}}}\mathbb{1}_{{0,\ldots,2n}}(x)$. Now $|g_n(x)|\leq (\sqrt[3]{2})^{-x}=g(x)$ and $\int g,d\nu <\infty$. Notice $\lim_ng_n(x)=2^{-x}$. – Mittens Aug 19 '21 at 22:55

1 Answers1

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Let $a_n:=\sum_{k=0}^{2n}2^{-k}$.

$$ |x_n-a_n|\le \sum_{k=0}^{2n}2^{-k}|\cos(\sqrt{k/n})-1| \le \sum_{k=0}^{2n}2^{-k}\sqrt{k/n} $$ $$ = \frac{1}{\sqrt{n}}\sum_{k=0}^{2n}2^{-k}\sqrt{k} \le \frac{1}{\sqrt{n}}\sum_{k=0}^{\infty}2^{-k}\sqrt{k} \to 0 ~ (n \to \infty). $$ As $a_n \to 2$ also $x_n \to 2$ as $n \to \infty$.

Gerd
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  • Thanks Gerd. That was rather simple. The second limit is probably a little harder. – Mittens Aug 21 '21 at 12:53
  • I think the second limit is a bit more technical, but the same idea should work by using Lipschitz continuity of $x \mapsto 2^x$ on bounded sets. – Gerd Aug 21 '21 at 13:08