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$\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$

any ideas?


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This was mentioned as an example of a series for which you need to know quite a bit about rational approximations to $\pi$ to prove convergence, in a comment to the following MO question. mathoverflow.net/questions/54758/… –  George Lowther Feb 9 '11 at 21:14
@George, thanks for your comment. I would be happy to hear the details if someone here knows them.... –  the L Feb 9 '11 at 23:27
I haven't seen a proof of this, but I think the result will follow from equidistribution, showing that the positive and negative terms in the series largely cancel each other. You will probably need to use the fact that $\pi$ has finite irrationality measure to get a good enough bound to prove convergence, as with David Speyer's response to this previous question: math.stackexchange.com/questions/2270/… –  George Lowther Feb 10 '11 at 1:09
If $f$ is a smooth function with period 1 and $\alpha$ has finite irrationality measure, then you can bound $\sum_{n=1}^Nf(\alpha n)-N\int_0^1f(x)dx$. Then, you can bound $\sum_{n=1}^N(-1)^nf(\alpha n)$. I think this could work for this sum by appying equidistribution to the $\cos n$ part, but I haven't been through the details yet. –  George Lowther Feb 10 '11 at 1:13

2 Answers 2

up vote 16 down vote accepted

This replaces my original answer, which was faulty but led in the right direction (I think).

I'll let the series start at some $n_0\ge2$; the reason for this will become apparent later.

Expand in $\cos(n)/\ln(n)$ (with $\lvert\cos(n)/\ln(n)\rvert < 1$ for $n \ge 2$):

\[\sum_{n=n_0}^\infty \sum_{k=0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]

Let's interchange the summations and worry later whether that was OK:

\[\sum_{k=0}^\infty \sum_{n=n_0}^\infty \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]

Now the inner sum converges according to Dirichlet's test. In fact we can proceed as in the proof of Dirichlet's test to bound its limit (suppressing indices $k$ while we work at fixed $k$):

\[a_n=\ln^{-k-1}(n)\] \[b_n=(-1)^n\cos^k(n)=\sum_{j=0}^k\alpha_{jk}\cos((j + \pi)n)\;,\]

with $\lvert\alpha_{jk}\rvert \le 1$, using the power-reduction formula for the cosine and absorbing the factor $(-1)^n$ into the cosines by adding $\pi$ to their frequency.

\[B_m=\sum_{n=n_0}^m \; b_n\]

\[\lvert B_m\rvert = \left\lvert\sum_{n=n_0}^m \sum_{j=0}^k \alpha_{jk}\cos((j + \pi)n)\right\rvert \le \sum_{j=0}^k\left\lvert\sum_{n=n_0}^m\cos((j + \pi)n)\right\rvert \le \sum_{j=0}^k\frac{2}{1 - \cos(j + \pi)}=:M_k\]

\[\left\lvert\sum_{n=n_0}^m a_n b_n\right\rvert = \left\lvert B_m a_{m+1}+\sum_{n=n_0}^{m}B_n(a_n-a_{n+1})\right\rvert\le\] \[\le M_k a_{m+1} + \sum_{n=n_0}^{m}M_k(a_n-a_{n+1})= M_k a_{m+1} + M_k(a_{n_0}-a_{m+1}) = M_ka_{n_0}\]

Thus, we can apply the comparison test with the following series:

\[\sum_{k=0}^\infty M_k\ln^{-k-1}(n_0)\]

This shows why we need to start at some $n_0$ instead of $2$, which wouldn't work since $\ln 2 < 1$.

Now everything depends on the behaviour of $M_k$. The cosine comes arbitrarily close to $1$; when $j$ comes within $\epsilon$ of an odd multiple of $\pi$, then $1/(1-\cos (j + \pi)) \approx 1/(1-(1 - \epsilon^2/2))= 2/\epsilon^2$. Heuristically speaking, we might expect this to happen every $1/\epsilon$ integers, so on average these spikes would not destroy the long-term decay with $\ln^{-k-1}(n_0)$ (which can be made quantitively, though not qualitatively, stronger by increasing $n_0$). However, I don't know how to make this argument rigorous, or whether that is even possible given the "unpredictability" of $\pi$. It's interesting that this proof might depend on the details of $\pi$ in this way. [Update: Thanks to George Lowther for the comments pointing to this MathWorld page, which gives an upper bound on the irrationality measure of $\pi$ (never heard of that before :-). That this is finite implies that $M_k$ can be bounded by a power of $k$, and hence the outer sum converges (uniformly) -- I believe that completes the proof.]

Assuming that the spike problem can be solved, we still have the interchangeability of the sums to come back to. This theorem shows that we may interchange the sums if the convergence of

\[\sum_{k=0}^\infty \sum_{n=n_0}^m \frac{(-1)^n}{\ln(n)} \frac{\cos^k(n)}{\ln^k(n)}\]

is uniform in $m$ -- which it is, since $m$ dropped out in the derivation of the bound $M_ka_{n_0}$.

To summarize, this derivation shows that the series can be shown to converge if the irregular spikes in $1/(1-\cos (j + \pi))$ that occur because $j$ gets arbitrarily close to odd multiples of $\pi$ can be shown not to destroy the convergence of the outer sum over $k$.

Thanks for an interesting question :-).

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thank your for a nice answer! –  the L Feb 10 '11 at 15:14
I don't follow how you conclude the last inequality in the line that starts $|B_m|.$ Putting a few random figures into wolframalpha gives $\sum_{n=5}^{100} \cos((1+\pi)n) \approx 0.689574$ but $1/(1-\cos(1+\pi)) \approx 0.649223,$ which appears to contradict what you've written when $j=1.$ –  Derek Jennings Feb 10 '11 at 16:33
@Derek Jennings: I'll look into that later tonight; I may have lost a factor of 2 somewhere, but the general idea is to write the cosine as a sum of exponentials and use the formula for the partial sum of a geometric series, so I think the inequality is right except perhaps for a wrong factor. More later... –  joriki Feb 10 '11 at 17:39
@Derek Jennings: You're right, I dropped a factor of 2. Thanks for spotting that. If you reduce the two fractions you get from the geometric series to a common denominator, you get 8 terms of absolute value $1$ in the numerator, and $4(1-\cos(j+\pi))$ in the numerator, so the bound has to be $2 / (1-\cos(j+\pi))$. –  joriki Feb 10 '11 at 17:55
@joriki: I mean that there actually are values of $\alpha$ for which it diverges. If $\alpha/\pi=p/q$ is rational with odd $p,q$, then it can be seen that $\cos \alpha n$ is periodic and the odd and even terms in the series do not quite balance. So, it diverges. Even irrational $\alpha$ can lead to divergence if $\alpha/\pi$ can be approximated well enough by odd fractions. –  George Lowther Feb 11 '11 at 22:15

Inserting \sum_{n=2}^{infty} (-1)^n/(ln(n)+cos(n)) into the search box of Wolfram Alpha gives an approximate value of –4.552162944044021. However unbelievable the actual sum may be, you can be a bit sure about its convergence.

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thanks for the comment, but I was hoping for a proof. –  the L Feb 9 '11 at 19:31

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