It is well known that $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent while $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon}}$ is convergent for a fixed positive value of $\epsilon$.

It is not difficult to show that $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$ is divergent using Limit comparison test with $ \displaystyle\frac{1}{n}$. There is a post on this question here.

Now comes my questions:

(i) Is $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+\left|{\cos n}\right|}}$ convergent or divergent? (I have tried several tests, like: comparison/limit comparison tests, but fail to get conclusion. My intuition is that it is divergent...)

(ii) It was stated here that $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2-\cos n}}=\sum_{n=1}^{\infty} \frac{1}{n^{1+(1-\cos n)}}$ is divergent. So is there is general way to determine if $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+f(n)}}$ with $f(n)>0$ for all natural number $n$, a convergent or divergent series?

Any comment or answer?

  • 1
    $\begingroup$ If you replace cosine with sine, the answer is here: math.stackexchange.com/questions/270064/… $\endgroup$
    – user940
    Feb 22, 2013 at 1:06
  • $\begingroup$ @ByronSchmuland Thanks! From the link provided, some post mentioned similar questions... $\endgroup$
    – pipi
    Feb 23, 2013 at 2:25
  • $\begingroup$ I am just a high school student and would like to learn whether we can compare it with a p series. What I mean is that $|cos(n)|$ is always positive, so the power of $n$ is larger than $1$, which indicates that the series is convergent. What am I missing? $\endgroup$
    – Mathrix
    Jan 9, 2020 at 23:45
  • $\begingroup$ @Mathrix what value of $p$ would you use? $1+|\cos(x)|$ can be arbitrarily close to $1$, so there is no single value of $p$ that can be used. In my answer, I use the uniform density of $n$ mod $\pi$ to compute how often $p$ is a certain distance from $1$. It turns out that $p$ is too close to $1$ too often to allow the series to converge. $\endgroup$
    – robjohn
    Jul 29, 2020 at 16:36

1 Answer 1


Almost Convergent

If we assume that $n$ mod $\pi$ is equidistributed in $[0,\pi)$, then $$ \sum_{n=1}^\infty\frac1{n^{1+|\!\cos(n)|}}\tag1 $$ should converge when $$ \begin{align} \sum_{n=1}^\infty\frac2\pi\int_0^{\pi/2}n^{-1-\cos(x)}\,\mathrm{d}x &=\sum_{n=1}^\infty\frac2{n\pi}\int_0^1n^{-x}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag2\\ &=\sum_{n=1}^\infty\frac2{n\pi}\int_0^1e^{-x\log(n)}\left(1+O\!\left(x^2\right)\right)\mathrm{d}x\tag3\\ &=\sum_{n=1}^\infty\frac2{n\pi}\left(\frac{1-\frac1n}{\log(n)}+O\!\left(\frac1{\log(n)^3}\right)\right)\tag4 \end{align} $$ converges.

However, $\sum\limits_{n=1}^\infty\frac1{n\log(n)}$ diverges (just barely). Therefore, this reasoning says that $(1)$ should also diverge.

Less Convergent

Still assuming the equidistribution of $n$ mod $\pi$ in $[0,\pi)$, $$ \sum_{n=1}^\infty\frac1{n^{2-\cos(n)}}\tag5 $$ should converge when $$ \begin{align} \sum_{n=1}^\infty\frac2\pi\int_0^{\pi/2}n^{-2+\cos(x)}\,\mathrm{d}x &=\sum_{n=1}^\infty\frac2\pi\int_0^1n^{-2+x}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag6\\ &=\sum_{n=1}^\infty\frac2\pi\int_0^1n^{-1-x}\frac{\mathrm{d}x}{\sqrt{2x-x^2}}\tag7\\ &=\sum_{n=1}^\infty\frac{\sqrt2}{n\pi}\int_0^1e^{-x\log(n)}\left(x^{-1/2}+O\!\left(x^{1/2}\right)\right)\mathrm{d}x\tag8\\ &=\sum_{n=1}^\infty\frac{\sqrt2}{n\pi}\left(\frac{\sqrt\pi}{\log(n)^{1/2}}+O\!\left(\frac1{\log(n)^{3/2}}\right)\right)\tag9 \end{align} $$ converges.

However, $\sum\limits_{n=1}^\infty\frac1{n\log(n)^{1/2}}$ diverges (faster than $(4)$). Therefore, this reasoning says that $(5)$ should also diverge.

  • $\begingroup$ Sorry for being a n00b, but how reasonable is the assumption $n$ equidistributed mod $\pi$ in $[0,\pi)$? It seems like one of those plausible statements that's probably true but I have no idea if it is or how difficult it would be to prove. $\endgroup$
    – Integrand
    Jul 29, 2020 at 20:36
  • 1
    $\begingroup$ That can be shown using an argument similar to the pigeonhole argument of this answer. What is certain is that $$\lim_{n\to\infty}\frac{{\large|}[a,b]\cap\{k\text{ mod }\pi:1\le k\le n\}{\large|}}n=\frac{b-a}\pi$$ $\endgroup$
    – robjohn
    Jul 29, 2020 at 23:08

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