I want to apply the alternating series test to prove the convergence of the series $\sum_{n\geq 1} (-1)^n u_n$ where $u_1=1$ and $\forall n\in \mathbb{N},\quad u_{n+1}=\frac{\cos(u_n)}{n^\alpha}$ where $\alpha>0$.

We can show that $0\leq u_n\leq 1$ for all $n$ and that $u_n\to 0$. Numerically, it seems that the sequence $u_n$ is decreasing after a certain point, but I don't see how to prove it.

  • $\begingroup$ $\log ~\left[ \dfrac{1}{n^\alpha} ~\right] = -\alpha\log(n).~$ What happens to this logarithm, as $~n \to \infty~?~$ Also, the function $~\cos(u_n),~$ is a bounded function, that beyond a certain point, must always be positive. $\endgroup$ Commented May 25 at 15:31
  • $\begingroup$ $-\alpha \ln(n)\to -\infty$, but I don't see what you're trying to tell me $\endgroup$
    – Noname
    Commented May 25 at 15:55
  • $\begingroup$ I am sorry, but MathSE protocol, as I interpret it, does not permit me to say more. Since your posted question has been upvoted, rather than downvoted, apparently, I have the minority view. However, I judge your posted question to be of low quality, with respect to MathSE protocol. I was willing to provide a subtle hint, which I suspected might lead you on the right track to showing significant work on your own. Howevr, I have taken the situation as far as I think is appropriate. ...see next comment $\endgroup$ Commented May 25 at 18:04
  • 1
    $\begingroup$ With respect to this article on MathSE protocol, your posted question is missing much information. One glaring oversight is the double question of [1] Is the series, beyond a specific point, actually an alternating series ? and [2] Is the limit, as $~n \to \infty,~$ of $~|u_n|,~$ going to $~0~?~$ Do you see no relevance in my comment, with respect to these issues? For example, what is $~\displaystyle \lim_{r \to -\infty} e^r ~?~$ $\endgroup$ Commented May 25 at 18:09

1 Answer 1


The next observation comes handy:

Lemma. There exists $N, C, \beta > 1$ such that

$$ \left| u_{n-1} - u_n \right| \leq \frac{C}{n^\beta}$$

whenever $n \geq N$.

Proof. We first assume $n \geq 3$. Then by the recurrence relation and the mean value theorem,

\begin{align*} \left| u_n - u_{n-1} \right| &= \left| \frac{\cos(u_{n-1})}{(n-1)^{\alpha}} - \frac{\cos(u_{n-2})}{(n-2)^{\alpha}} \right| \\ &\leq \left| \frac{\cos(u_{n-1}) - \cos(u_{n-2})}{(n-1)^{\alpha}} \right| + \left| \frac{1}{(n-1)^{\alpha}} - \frac{1}{(n-2)^{\alpha}} \right| \left| \cos(u_{n-1}) \right| \\ &\leq \frac{\left| u_{n-1} - u_{n-2} \right|}{(n-1)^{\alpha}} + \frac{\alpha}{(n-2)^{\alpha + 1}}. \end{align*}

Repeatedly applying this inequality as many (but fixed) number of times, the desired conclusion follows. $\square$

Now we return to the original claim. Let $N, C, \beta$ be as in the lemma. By increasing $N$ if needed, we may assume $N^{\beta - 1} \geq \frac{C\tan 1}{\alpha}$. Now assume $0 < u_n \leq u_{n-1}$ for some $n \geq N$. Then by noting that $|u_n| \leq 1$,

\begin{align*} \log u_{n+1} - \log u_n &= \log\cos(u_n) - \log\cos(u_{n-1}) + \alpha \log\left(1 - \frac{1}{n}\right) \\ &= \int_{u_n}^{u_{n-1}} \tan x \, \mathrm{d}x - \alpha \int_{0}^{\frac{1}{n}} \frac{\mathrm{d}x}{1 - x} \\ &\leq (u_{n-1} - u_n) \tan 1 - \frac{\alpha}{n} \\ &\leq \frac{C \tan 1}{n^\beta} - \frac{\alpha}{n} \leq 0 \end{align*}

by our choice of $N$. In particular, this implies that $u_{n+1} \leq u_n$. Then by the principle of mathematical induction, we have $u_{k+1} \leq u_k$ for all $k \geq n$. This observation and mathematical induction together, we find that:

Observation. If $0 < u_n \leq u_{n-1}$ holds for infinitely many $n$, then $0 < u_n \leq u_{n-1}$ for all sufficiently large $n$ (and hence $u_n$ is eventually non-increasing.)

However, the condition of this observation is clearly satisfied since $u_n > 0$ and $u_n \to 0$.

  • $\begingroup$ Thank you for this comprehensive response. I need some time to study it. $\endgroup$
    – Noname
    Commented May 25 at 18:17
  • $\begingroup$ Thank you very much, I was not familiar with this kind of recurrence $\endgroup$
    – Noname
    Commented May 25 at 20:09

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