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Help prove the alternating series $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n}$ is convergent. $\frac 1 {\ln n + \sin n}$ is a decreasing sequence but it is not motonically decreasing. I am not sure how to deal with this situation.

My failed attempt..

For even terms,

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

where the two "bound" series do not converge

For odd terms

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

where the two "bound" series do not converge.

Community
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user432594
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    Try using the fact that $-1\leq \sin(n)\leq 1$. – DMcMor Apr 05 '17 at 00:01
  • By itself, those inequalities don't help: the Alternating Series Test has hypotheses that are false for this series. – Greg Martin Apr 05 '17 at 00:07
  • The alternating series only requires that $a_n$ is eventually decreasing. A finite amount of terms does not effect convergence. – Ahmed S. Attaalla Apr 05 '17 at 00:08
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    The function $\frac1{\ln x+\sin x}$ is not eventually decreasing. – Greg Martin Apr 05 '17 at 00:09
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    First time I've seen "{\text{ + }}" in MathJax code where simply "+" was appropriate. – Michael Hardy Apr 05 '17 at 00:35
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    This is pretty hardcore. I don't think this will be any easier than http://math.stackexchange.com/questions/21175/does-the-following-series-converge (which is the same series, only with $\sin(n)$ replaced by $\cos(n)$). – levap Apr 05 '17 at 00:52
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    In the @levap's link, the series with $\cos(n)$ converges. The same method applies to this one. Thus, this series converges. – Sungjin Kim Apr 05 '17 at 23:17
  • @i707107 The argument there does not show that the series converges. Rather, it reduces the problem to an open question in what is in essence Diophantine approximations. – Andrés E. Caicedo Jun 07 '17 at 00:35
  • @AndrésE.Caicedo In the link, the answer was edited at least once. First version without the knowledge on Diophantine approximations, and second version after the Diophantine approximation result provided by George Lowther. – Sungjin Kim Jun 07 '17 at 00:50
  • @AndrésE.Caicedo Also, finiteness of irrationality measure of $\pi$ is not an open question. See http://mathworld.wolfram.com/IrrationalityMeasure.html – Sungjin Kim Jun 07 '17 at 00:51
  • @i707107 Ah, yes. Nice. Thanks. – Andrés E. Caicedo Jun 07 '17 at 00:52
  • @i707107 Yes, I first only skimmed the answer and thought that a more precise estimate was needed. Nice problem. – Andrés E. Caicedo Jun 07 '17 at 00:58
  • @AndrésE.Caicedo Although finiteness of irrationality measure of $\pi$ is known, but finding an exact value of it is still an open question. Some infinite series such as $\sum 1/(n^3 \sin^2 n)$ depends heavily on the value of it. This series is known as Flint Hills series, and it is not known whether it converges or diverges. – Sungjin Kim Jun 07 '17 at 02:59

1 Answers1

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Does $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n} $ converge or diverge?

I think that it probably diverges, but I can show that the similar sum $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + (-1)^n} $ diverges and $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + c} $ converges for any real $c$ such that $\ln(n)+c \ne 0$ for all $n$ (in particular for $c > 0$).

If $f(n)$ is bounded, let $s_m =\sum\limits_{n = 1}^m \frac{(-1)^n}{\ln n + f(n)} $ and let $s =\lim_{m \to \infty} s_m $ if the limit exists.

$\begin{array}\\ s_{2m} &=\sum\limits_{n = 1}^{2m} \frac{(-1)^n}{\ln n + f(n)}\\ &=\sum\limits_{n = 1}^{m} (\frac{-1}{\ln (2n-1) + f(2n-1)}+\frac{1}{\ln (2n) + f(2n)})\\ &=\sum\limits_{n = 1}^{m} \frac{-(\ln (2n) + f(2n))+(\ln (2n-1) + f(2n-1))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n)) + f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}+\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\ &=u_m+v_m\\ \end{array} $

where $u_m =\sum\limits_{n = 1}^{2m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))} $ and $v_m =\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))} $.

Each term in $u_m$ is, for large enoough $n$, less in absolute value than $\frac1{n\ln^2(n)} $ (because $\ln(1-1/(2n)) \approx -\frac1{2n}$ and $(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n)) \approx \frac1{\ln^2(n)} $) and the sum of these converges.

Therefore convergence depends on $v_m$. Each term in $v_m$ is $ \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))} \approx \frac{f(2n-1)-f(2n)}{\ln^2(2n)} $.

If $f(n) = c$, where $c$ is a constant, then $f(2n-1)-f(2n) = 0$, so $v_m = 0$ and the sum converges.

If $f(n) = (-1)^n$, then $f(2n-1)-f(2n) = (-1)-(1) = -2 $ and the sum of $\frac{-2}{\ln^2(n)} $ diverges.

If $f(n) = \sin(n)$,

$\begin{array}\\ f(2n-1)-f(2n) &=\sin(2n-1)-\sin(2n)\\ &=2\sin((2n-1-2n)/2)\cos((2n-1+2n)/2)\\ &=2\sin(-1/2)\cos(2n-1/2)\\ \end{array} $

so the sum depends on the sum $\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)} $.

According to Wolfy, $\sum_{n=1}^m \cos(2 n - 1/2) = \csc(1) \sin(m) \cos(m + 1/2) $ so this sum is bounded but not convergent. I think that this implies that $\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)} $ does not converge. This might be proved using a variation of summation by parts, but I am not sure how to do this, so I will leave it at this.

marty cohen
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  • Good job, I was starting to suspect this sum diverges, just because nobody can manage to prove it converges. But I'd still love to see a complete rigorous proof. – Gregory Grant Jul 26 '17 at 20:14
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    Your sums after the first one should be to $m$ not to $2m$. Also how do you get the bound on $u_m$ as less than $\frac1{n\ln^2(n)}$ for large enough $n$? – Gregory Grant Jul 26 '17 at 23:30
  • Thanks. I made the correction and added an additional explanation. – marty cohen Jul 27 '17 at 04:04
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    The integral $$ \int_1^\infty \frac{\cos(2t-1/2)}{\ln^2(2t)} ,dt $$ does converge (use integration by parts, differentiating $\frac1{\ln^2(2t)}$ and integrating $\cos(2t-1/2)$). So I suspect the sum $$\sum_{n=1}^\infty \frac{\cos(2n-1/2)}{\ln^2(2n)}$$ also converges, and indeed summation by parts is probably how it can be proved. – Greg Martin Jul 27 '17 at 04:31
  • What I consider probable is that the sum is bounded but oscillates, so that it does not converge to a limit. – marty cohen Jul 27 '17 at 04:43
  • The series $\sum_{n=1}^\infty \frac{\cos(2n-1/2)}{\ln^2(2n)}$ is convergent by the Dirichlet test since $1/\ln^2(2n) \downarrow 0$ and you already showed the partial sums of $\cos(2n-1/2)$ are bounded. – RRL Jul 27 '17 at 07:53
  • @martycohen In wolfram says that the sum converges and it gives also a value of the sum. – Marios Gretsas Jul 27 '17 at 10:21
  • What value? Alpha seems to show the partial sums oscillating from about -.13 to -.15. – marty cohen Jul 27 '17 at 15:26
  • The sum converges. See https://math.stackexchange.com/questions/21175/convergence-of-the-series-sum-limits-n-2-infty-frac-1n-lnn?noredirect=1&lq=1, also https://math.stackexchange.com/questions/2384464/does-the-series-sum-limits-n-2-infty-1n-ln-left1-frac-sin-n-ln-n-r. The same method applies. – Sungjin Kim Aug 08 '17 at 23:12