# Determining convergence or divergence of series $\sum_{n=1}^{\infty} \frac{(-1)^n}{ n+\sin (n)}$

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...



\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.
• the last series "converges absolutely" // Is this right 'comparison test with 1/n(n-1)' ? // BTW I am impressed by your answer, thank you – user143993 Sep 27 '14 at 22:44
• Yes, the terms are bounded in absolute value by $1/(n-1)n$. – Julián Aguirre Sep 28 '14 at 15:34
• // THANK YOU VERY MUCH. Have a good day. – 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,

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

• your idea "add two consecutive terms together".. BUT SUM (-1)^n diverges.. // could you show appropriateness of your idea, please? // THANK YOU – user143993 Sep 27 '14 at 22:33
• 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. – Rene Schipperus Sep 27 '14 at 22:58

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

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

• 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 – user143993 Sep 27 '14 at 22:37
• this is called squeeze lemma – Alex Sep 27 '14 at 22:41
• 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 – user143993 Sep 27 '14 at 22:59
• both upper and lower series converge, hence $\sum c_n$ converges too – Alex Sep 27 '14 at 23:03
• Also in this case $c_n \sim a_n \sim b_n$ and both $\sum a_n$ and $\sum b_n$ converge, hence... – Alex Sep 27 '14 at 23:04