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Let us consider series:

$$S=\sum_{n=1}^{\infty} a_n$$

$$a_n=\frac{n\cos(n)+\sin(n^2+2n-5)}{n^2}$$

The bounds for $a_n$ is shown in this MSE question to be:

$$|a_n| \le \frac{n+1}{n^2}$$

Is it possible to prove that this series is convergent?

Numerical results showed that $$\sum_{n=1}^{10^5} a_n=-0.931506$$ $$\sum_{n=1}^{10^6} a_n=-0.931501$$

Here is a plot of Christian Blatter's bounds $\pm (n+1)/n^2$ and $a_n$: enter image description here

From the plot the series, we can see that the terms are oscillating, but not like $(-1)^n |a_n| $. There seems to be some randomness in the number of terms to be positive and number of terms to be negative.

Community
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mike
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    So it is an alternating series? This should be specified. Otherwise it would diverge – graydad Sep 04 '14 at 13:44
  • alternating property is what needs to be proved. – mike Sep 04 '14 at 13:59
  • "rom the plot the series, we can see that the terms are alternating" Actually they are not. Recall that "alternating" means that the signs are $+-+-+\ldots$. – Did Sep 04 '14 at 14:05
  • @Did. Thanks. I changed "alternating" to "oscillating". – mike Sep 04 '14 at 14:20

1 Answers1

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  • $$|\frac{\sin(n^2+2n-5)}{n^2}|\leq\frac{1}{n^2}$$

  • $\sum\frac{\cos n}{n}$ is convergent by using dirichlet test.

gaoxinge
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  • gaoxinge nice answer! – mike Sep 04 '14 at 13:58
  • Maybe I'm thick, but is it easy to see that the sum $\sum^N_{k=0}\cos(k)$ is bounded uniformly in $N$? I assume by 'Dirichlet test' you're reffering to this – Teri Sep 04 '14 at 14:01
  • @Teri This definitely can be done but the real question is whether the OP (mike) can do it. – Did Sep 04 '14 at 14:03
  • @Did. You are right. I can not do it. I took some math classes in physics dept when I was a freshman long time ago and can not remember if we were taught about "Dirichlet test" for convergence. ;) – mike Sep 04 '14 at 14:09
  • Ahha! I was indeed thick; write $\cos (k) =1/2 \ (e^{ik}+e^{-ik})$ and use the formula for the partial sums of the geometric series. If mike is aqcuainted with these, he (at least now) can do it. – Teri Sep 04 '14 at 14:10
  • Mathematica showed that $\sum_{n=1}^{\infty} (1/n)\cos n=-(1/2)\log(2(1-\cos 1))$ – mike Sep 04 '14 at 14:15
  • "I can not do it." But you accepted the answer? I will never get it. – Did Sep 04 '14 at 15:35