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How to prove this serie

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

converge? I can't do a comparison test with the Leibniz formula for $\pi$ because the series are not $>0$ for all $n$. I can't do a ratio test because I can't compute the limit, the alternating series test can't be applied, the absolute serie is not convergent. I'm out of ideas.

Any clues?

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    $\begingroup$ See comments here (in particular the link to the other question with $\cos $) math.stackexchange.com/questions/2218454/… $\endgroup$ – Zain Patel Jun 3 '17 at 19:06
  • $\begingroup$ @ZainPatel Thanks, I tried searching for the same question here but couldn't find it. Sorry $\endgroup$ – lombardo2 Jun 3 '17 at 19:08
  • $\begingroup$ pesky $\endgroup$ – Zubin Mukerjee Jun 3 '17 at 19:55
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Let $$a_n = \frac{(-1)^n}{\sin n + \ln n }$$

We wish to show that $\sum a_n$ converges.

In order to do this, first note that $a_n$ is negative when $n$ is odd, and positive when $n$ is even.

We will write $\sum a_n < \sum b_i + \sum c_j$, where $b_i$ are negative terms (with only odd indices $i$) that are smaller in absolute value than the negative terms $a_i$, and $c_j$ are positive terms (with only even indices $j$) that are larger in absolute value than the positive terms $a_j$.


We want to choose $\{b_i\}$ and $\{c_j\}$ to satisfy the following conditions:

  • $\sum b_i$ (sum taken over odd $i$) converges by the alternating series test

  • $\sum c_j$ (sum taken over even $j$) converges by the alternating series test

  • $|a_i| > |b_i|$ or equivalently, $a_i < b_i$, for all odd $i$

  • $a_j < c_j$ for all even $j$


For $i$ odd, let $$b_i = \frac{-1}{2 + \ln n } > a_i$$

For $j$ even, let $$c_j = \frac{1}{-2 + \ln n } > a_i$$

(I am choosing $2$ and $-2$ here because they are greater than the maximum of $\sin n$ and less than the minimum of $\sin n$, respectively)


Note that $b_i$ and $c_j$ are both monotonically decreasing for $i,j > 10$.

(I am choosing $10$ here because it is greater than $e^2$, to avoid negative denominators due to $-2 + \ln n$)

Therefore, by the alternating series test, we know that the following sums must converge:

$$\displaystyle\sum_{i=11,\, i \text{ odd}}^\infty b_i$$

$$\displaystyle\sum_{j=10,\, j \text{ even}}^\infty c_j$$


Therefore, $\sum a_n$ is bounded above by the sum of two convergent series: $$\displaystyle\sum_{n=10}^\infty a_n < \left(\displaystyle\sum_{i=11,\, i \text{ odd}}^\infty b_i \right) + \left(\displaystyle\sum_{j=10,\, j \text{ even}}^\infty c_j\right)$$

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    $\begingroup$ the absolute value of the series is not monotone. The alternating series test requires the absolute values be monotonically decreasing. $\endgroup$ – robjohn Jun 3 '17 at 19:00
  • $\begingroup$ Don't I need the absolute values of the terms to be monotonically decreasing? That's the first test I tried, what am I missing? $\endgroup$ – lombardo2 Jun 3 '17 at 19:02
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    $\begingroup$ How does convergence of the new series yield convergence of the original series ? $\endgroup$ – Gabriel Romon Jun 3 '17 at 19:13
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    $\begingroup$ Replace $\sin n$ with $(-1)^n / n$. This is still bounded below by $-1$ so your argument applies equally. Hence $\sum (-1)^n / (\ln n + (-1)^n / n)$ converges by your argument. $\endgroup$ – Zain Patel Jun 3 '17 at 19:17
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    $\begingroup$ I've made another edit - I am still not completely sure about it - I can show that $\sum a_n$ is bounded above by the sum of two convergent series, and bounded below by the sum of two different convergent series, but this still doesn't technically prove convergence, does it? Because the sequence could oscillate forever somewhere between those two bounds - intuitively it cannot, since the terms go to zero in absolute value ... This problem can definitely be solved using the alternating series test - or at least the idea behind it .... but making it rigorous is painful :( $\endgroup$ – Zubin Mukerjee Jun 3 '17 at 19:48

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