Solution to Problem 7

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Solution:

It is quite easy to see that for any rational \(x=\frac{p}{q}, q\neq0\) and \(q>0\), \(n!\frac{p}{q}\) turns out to be an even number for \(n> p+1\). Thus, \(\cos(n!\pi x)\) is simply \(1\) for all \(n>p+1\). Thus the sequence converges to \(1\) for \(x\) rational. Now, for the irrational number \(2e\) the series converges to \(1\) and hence for any number of the form \(2e+2k\) the sequence converges, where \(k\) is an integer.

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Abhishek M. N. (M.Sc. student, Manasagangothri), Nagendra P, (Research Scholar, Manasagangotrhi) and Manjunath M. R. (Reasearch scholar, IIT Indore) gave correct answers for the rational part. However, although Nagendra and Manjunath guessed that for \(e\) the sequence may converge to \(1\), they could not give a full proof.

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Solution for the irrational part We know that

\(2e = 2+\frac{2}{1!}+\frac{2}{2!}+\frac{2}{3!}+\dots+\frac{2}{k!}+\dots\) Thus \(n!2e = \underbrace{2n!+\frac{2n!}{1!}+\frac{2n!}{2!}+\frac{2n!}{3!}+\dots+\frac{2n!}{n!}}_{\text{Note that this is an even integer say \(2k\)}}+\frac{2n!}{(n+1)!}+\frac{2n!}{(n+2)!}+\dots\) Hence \(n!2e = 2k +\frac{2}{n+1}+\frac{2}{(n+1)(n+2)}+\frac{2}{(n+1)(n+2)(n+3)}+\dots+\frac{2}{(n+1)(n+2)\dots(n+k)}+\dots \) Thus \( \cos(n!2e\pi) = \cos\left(\frac{2\pi}{(n+1)(n+2)}+\frac{2\pi}{(n+1)(n+2)(n+3)}+\dots+\frac{2\pi}{(n+1)(n+2)\dots(n+k)}+\dots\right)\) Now, we shall show that as \(n\to \infty\) \( \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots+\frac{1}{(n+1)(n+2)\dots(n+k)}+\dots \to 0\). This is quite easy, as \(\frac{1}{(n+1)(n+2)\dots(n+k)} < \underbrace{\frac{1}{(n+1)(n+1)\dots(n+1)}}_{\text{\(k\) times}}=\frac{1}{(n+1)^k}\). Thus \(\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots+\frac{1}{(n+1)(n+2)\dots(n+k)}+\dots \leq \sum\limits_{k=1}^{\infty} \frac{1}{(n+1)^k}=\frac{1}{n} \). Thus by squeeze lemma, as \(n\to \infty\), we must have that \( \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots+\frac{1}{(n+1)(n+2)\dots(n+k)}+\dots \to 0\). Hence \(\lim\limits_{n\to \infty} \cos(n!\pi x) = \cos(0) = 1.\)