I am wondering the convergence or divergence following series

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

My 1st attempt is 'alternating series test'

$$ $$

But, $$\frac{1}{n+\sin(n)}$$

isn't monotone decreasing. SO I failed...

$$ $$

Please give me some advice.

Thanks in advance.


\begin{align*} \frac{(-1)^n}{n+\sin n}&=\frac{(-1)^n}{n}+\Bigl(\frac{(-1)^n}{n+\sin n}-\frac{(-1)^n}{n}\Bigr)\\ &=\frac{(-1)^n}{n}-\frac{(-1)^n\sin n}{(n+\sin n)n}. \end{align*}

  • $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges conditionally.
  • $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n\sin n}{(n+\sin n)n}$ converges absolutely.
  • $\begingroup$ the last series "converges absolutely" // Is this right 'comparison test with 1/n(n-1)' ? // BTW I am impressed by your answer, thank you $\endgroup$ – user143993 Sep 27 '14 at 22:44
  • $\begingroup$ Yes, the terms are bounded in absolute value by $1/(n-1)n$. $\endgroup$ – Julián Aguirre Sep 28 '14 at 15:34
  • $\begingroup$ // THANK YOU VERY MUCH. Have a good day. $\endgroup$ – user143993 Sep 30 '14 at 0:32

We have

$$\frac{(-1)^n}{n+\sin n}=\frac{(-1)^n}{n}\left(1+\frac{\sin n}{n}\right)^{-1}=\frac{(-1)^n}{n}\left(1-\frac{\sin n}{n}+o\left(\frac1n\right)\right)\\=\frac{(-1)^n}{n}+\frac{(-1)^{n+1}\sin n}{n^2}+o\left(\frac1{n^2}\right)$$ so the given series is convergent since it's sum of $3$ convergent series:

  • the first by the Leibniz theorem
  • the second the third by comparison with a Riemann series.

add two consecutive terms together,

$$\frac{1}{n+\sin n}-\frac{1}{n+1+\sin (n+1)}=\frac{\sin (n+1)-\sin n +1}{(n+\sin n)(n+1+\sin (n+1))}$$ Now do a compairison,

$$\frac{|\sin (n+1)-\sin n +1|}{(n+\sin n)(n+1+\sin (n+1))}\leq \frac{3}{(n-1)n}$$

  • $\begingroup$ your idea "add two consecutive terms together".. BUT SUM (-1)^n diverges.. // could you show appropriateness of your idea, please? // THANK YOU $\endgroup$ – user143993 Sep 27 '14 at 22:33
  • $\begingroup$ The terms of $(-1)^n$ do not limit to zero. I am showing that the even partial sums converge, but as they differ, in this case, from the odd partial sums by a quantity that limits to zero, the odd sums also converge and to the same sum. $\endgroup$ – Rene Schipperus Sep 27 '14 at 22:58

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I am not very sure that my answer is correct or not as these tests are what i learned. Hope this may help you.

  • $\begingroup$ This is wrong as you show divergence of absolute, but not alternating series. This alternating series converges by Leibniz test $\endgroup$ – Alex Sep 27 '14 at 15:43
  • $\begingroup$ @Alex you are right...So this mean that this series converge conditionally right? $\endgroup$ – Alan Wang Sep 27 '14 at 15:52
  • $\begingroup$ Yes, check Wikipedia for alternating series test $\endgroup$ – Alex Sep 27 '14 at 15:53

Easier: see that $-1 \leq \sin n \leq 1 $, hence the bounds on the summand are $\frac{(-1)^n}{n-1}$ and $\frac{(-1)^n}{n+1}$ which converge by alternating test (Leibniz test) and hence your original series converges.

  • $\begingroup$ In case of "a_n < c_n < b_n" and SUM a_n = α, SUM b_n = β ,., HOW can I say SUM c_n is converges? // Would you please tell me supplement? // Thank you $\endgroup$ – user143993 Sep 27 '14 at 22:37
  • $\begingroup$ this is called squeeze lemma $\endgroup$ – Alex Sep 27 '14 at 22:41
  • $\begingroup$ I know squeeze thm for limit and for case an<bn<cn in addition an, cn are converges SAME limit...// BUT this case.. wait a minute, I'll find squeeze thm for series, THANK YOU very much $\endgroup$ – user143993 Sep 27 '14 at 22:59
  • $\begingroup$ both upper and lower series converge, hence $\sum c_n$ converges too $\endgroup$ – Alex Sep 27 '14 at 23:03
  • $\begingroup$ Also in this case $c_n \sim a_n \sim b_n$ and both $\sum a_n$ and $\sum b_n$ converge, hence... $\endgroup$ – Alex Sep 27 '14 at 23:04

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