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How to prove that $$u_n=\sum_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$$ is a Cauchy sequence?

The exercise I am reading gives as a hint that we should use the inequality: $$0\leq\cos\left(\frac kn\right)^{2n^2/k}\leq e^{-k},$$ for all $k\leq n$.

I tried to estimate $|u_m-u_n|$, but I don't know how to deal with the $n$ inside the sum.

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    $\begingroup$ So writing out $u_3$, and then applying the hint to each term, we see that $u_3 \le e^{-1} + e^{-2} + e^{-3}$. Do you notice anything about the (short) series on the right? $\endgroup$ – John Hughes May 6 '18 at 0:54
  • $\begingroup$ Is there any reason that you have to show that it's a Cauchy sequence? Can you use the comparison test? $\endgroup$ – rwbogl May 6 '18 at 1:07
  • $\begingroup$ @JohnHughes This will give:$$0\leq \sum_{k=1}^n \cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}\leq \sum_{k=1}^n e^{-k}=\frac{e^{-1}-e^{-(n+1)}}{1-e^{-1}}.$$ But we can't conclude something about the convergence. $\endgroup$ – David Lingard May 6 '18 at 1:12
  • $\begingroup$ @rwbogl I guess they asked to prove it's Cauchy because we don't know the possible limit. $\endgroup$ – David Lingard May 6 '18 at 1:15
  • $\begingroup$ @DavidLingard Sure, but the comparison test doesn't require you to know the limit either. Using the hint to show that the partial sums are Cauchy is like proving the comparison test itself. $\endgroup$ – rwbogl May 6 '18 at 1:20
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Forget Cauchy sequences. To show the convergence of the sequence $(u_n)$, consider, for every integer positive $(k,n)$, $$v_{k,n}=\cos\left(\frac{k}{n}\right)^{2n^2/k}\mathbf 1_{k\leqslant n}$$ then $$u_n=\sum_{k=1}^\infty v_{k,n}$$ Assume momentarily that, for every $x$ in $[0,1]$, $$\cos x\leqslant e^{-x^2/2}\tag{$\ast$}$$ Then, for every positive $(k,n)$, $$|v_{k,n}|=v_{k,n}\leqslant\left(e^{-k^2/(2n^2)}\right)^{2n^2/k}=e^{-k}$$ and, for every fixed positive $k$, $$\lim_{n\to\infty}v_{k,n}=e^{-k}$$ Thus, by Lebesgue dominated convergence theorem for series, $$\lim_{n\to\infty} u_n=\sum_{k=1}^\infty\lim_{n\to\infty} v_{k,n}=\sum_{k=1}^\infty e^{-k}=\frac1{e-1}$$ To complete the proof, one needs to prove $(\ast)$ but this is direct, considering the derivative of the function $$f(x)=e^{x^2/2}\cos x$$ on $[0,1]$ and using the estimate $\tan x\geqslant x$ on this interval.

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  • $\begingroup$ Thanks, nice proof. So that means that $$\lim_{n\to \infty}\sum_{k=1}^\infty v_{k,n}$$ and $$\lim_{n\to \infty}\sum_{k=1}^n v_{k,n}$$ are the same thing? $\endgroup$ – David Lingard May 6 '18 at 15:16
  • $\begingroup$ Hmmm... please note that $v_{k,n}=0$ for every $k>n$ hence indeed, $$\sum_{k=1}^nv_{k,n}=\sum_{k=1}^\infty v_{k,n}$$ $\endgroup$ – Did May 6 '18 at 15:26
  • $\begingroup$ my bad, that's correct, thank you $\endgroup$ – David Lingard May 8 '18 at 22:10
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First of all, $\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}\leq e^{-k} $ is equivalent to $\cos\left(\frac{k}{n}\right) \leq e^{-k^2/2n^2} $ which follows from $e^{-x} \ge 1-x+x^2/2$ for $\frac12 \ge x \ge 0$ (so $e^{-k^2/2n^2} \ge 1-k^2/(4n^2)+k^4/(8n^4) $) and $\cos(x) \le 1-x^2/2+x^4/24$ (so $\cos(k/n) \le 1-k^2/(2n^2)+k^4/(24n^4) $). Note that we have to go to the $x^4$ term.

Then $u_n =\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \le\sum_{k=1}^n e^{-k} \lt \frac{1}{e-1} $ so $u_n$ is a bounded sequence.

However, we have not yet shown that $u_n$ is increasing.

Instead I will show that $u_n \to \frac{1}{e-1} $ with too much computation.

We have $\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \ge \left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k}} = \left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k $.

For $0 < x \le \frac12$,

$\begin{array}\\ -\ln(1-x) &=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\ &=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\ &\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\ &=x+\dfrac{x^2}{2(1-x)}\\ &\le x+x^2/4\\ \end{array} $

or $\ln(1-x) \ge -x-x^2/4 $.

Therefore, for $0 < x \le \frac12$, $(1-x)^{1/x} =\exp((1/x)\ln(1-x)) \ge\exp(-(1/x)(x+x^2/4)) =\exp(-1-x/4) $ so $\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}} \ge \exp(-1-k^2/(8n^2)) $ so $\left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k \ge \exp(-k-k^3/(8n^2)) $ so $u_n =\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \ge\sum_{k=1}^n \exp(-k)\exp(-k^3/(8n^2)) $.

I will now split the sum into two parts: $[1, n^c]$ and $(n^c, n]$ where $0 < c < 1$.

If $k > n^{c}$, all the terms are positive, so that sum is positive.

For the rest of $u_n$, since $e^{-x} \gt 1-x $,

$\begin{array}\\ \sum_{k=1}^{n^{c}} \exp(-k)\exp(-k^3/(8n^2)) &\gt \sum_{k=1}^{n^{c}} \exp(-k)(1-k^3/(8n^2))\\ &= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)k^3/(8n^2)\\ &= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)n^{3c-2}/8\\ &\gt \sum_{k=1}^{n^{c}} \exp(-k)-\frac{n^{3c-2}}{8}\sum_{k=1}^{n^{c}} \exp(-k)\\ \end{array} $

and the first term approaches $\sum_{k=1}^{\infty} \exp(-k) =\frac{1/e}{1-1/e} =\frac{1}{e-1} $ and the second term is less than $\frac{n^{3c-2}}{8(e-1)} $ which goes to zero if $c < \frac23$.

Therefore the sum approaches $\frac{1}{e-1}$.

Whew.

That was harder than I thought it would be.

Hope that it's correct.

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    $\begingroup$ Hi Marty. $\cos(x)$ can exceed $e^{-x^2/2}$. However, when $|x|<1$, the inequality is preserved. So, you might consider mentioning that the inequality of interest applies since $k\le n$. $\endgroup$ – Mark Viola May 6 '18 at 2:36
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Starting from the high school limit $\lim_{x\to 0}\log(\cos(x))/x^2=-1/2$, we have near $0$ that $\cos(x)\approx e^{-x^2/2}$. Intuitively (having in mind the Taylor series and focusing on the interval $x$ in $[0,1]$), it's clear now you can bound the sum using the exponential function as you state in your post.

More explanations: The high school limit gives us the starting idea, and the Taylor series of $e^{-x^2/2}-\cos(x)$, with $x$ in $[0,1]$ assures the desired inequality.

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  • $\begingroup$ @downvoter I didn't want to modify the post initially, but maybe it's time for you to learn. Show me where my mathematics fail. I only added the text in the brackets. $\endgroup$ – user 1591719 May 6 '18 at 19:40
  • $\begingroup$ This only gives us an approximation for $x\approx 0$. In the question, $x=\frac kn$ ranges from $0$ to $1$. $\endgroup$ – Akiva Weinberger May 6 '18 at 20:03
  • $\begingroup$ @AkivaWeinberger did you look at the Taylor series of $e^{-x^2/2}-\cos(x)$ combined with the fact that $x$ is in $[0,1]$? $\endgroup$ – user 1591719 May 6 '18 at 20:07
  • $\begingroup$ If that works, please edit your post with some details. $\endgroup$ – Akiva Weinberger May 6 '18 at 20:08
  • $\begingroup$ @AkivaWeinberger Is it crystal clear now? $\endgroup$ – user 1591719 May 6 '18 at 20:16
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By making use of the inequality $0\leq cos(\frac{k}{n})^{2n^2/k}\leq e^{-k}$ for all $k\leq n$ we can state that

$$\sum_{k=1}^n cos(\frac{k}{n})^{2n^2/k}\leq\sum_{k=1}^ne^{-k}$$

If an $f$ function verifies that $f'(x)<0$ and $f''(x)>0$ for all $x\in[u,v]$ then we can apply the following inequality

$$\sum_{k=u}^vf(k)\leq\int_u^{v+1}f(x)dx+f(u)-f(v+1)$$

As the function $e^{-x}$ meets the requirements of the statement for all $x\in\mathbb{R}$ we can prove

$$\sum_{k=1}^ne^{-k}\leq\int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$

If $n\in\mathbb{N}$ it is convergent, so we just have to know what happens when $n\rightarrow\infty$. Hence,

$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq\lim_{n\rightarrow\infty} \int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$

$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq -e^{-x}|_1^\infty+e^{-1}=\frac{2}{e}$$

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  • $\begingroup$ We can simply deduce using geometric sequences that $\sum_{k=1}^{\infty} e^{-k}=\frac{1}{e-1}$. But then again, this only shows that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ is bounded. $\endgroup$ – David Lingard May 8 '18 at 22:16
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Hint. You have at least two options:

  • use formulas for geometric series,
  • use a comparison with the integral $\int_{-\infty}^0e^{t}dt=1$
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