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I wanted to also ask about this one

the sum of (1 +sin^2(n))/n from n =1 to n=00. I tried the divergence test and got infinity. But the graph of the functions shows it converges to zero. Any suggestions?

Does the sequence $\left\{\frac{1}{n}\right\}_{n=1}^\infty$ converge or diverge?

The same question for the sequence $\left\{\arctan(n)\right\}_{n=1}^\infty$. I know how to prove it diverges when it is a series. I used the divergence test. But when it is a sequence, how do I deal with it?

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    $\begingroup$ To prove a sequence converges, a common technique is to prove it's either bounded below and decreasing, or bounded above and increasing. $\endgroup$ – vrugtehagel May 6 '17 at 15:40
  • $\begingroup$ A sequence $(x_n)$ converges to $a\in\mathbb{R}$ if $\forall\epsilon>0,\exists N\in\mathbb{N}\ \text{such that} \forall n>N,|x_n-a|<\epsilon.$ Use this to prove the convergence. $\endgroup$ – Janitha357 May 6 '17 at 15:42
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The sequence with terms $1/n$ converges with limit 0. This is not hard to show. Use the definition of limit as suggested by @Janitha357. The sequence with terms $1+\sin^2(n)$ is bounded between 1 and 2 (Why?). So, the sequence you are interested in lies term-by-term between $\{1/n\}$ and $\{2/n\}$. As each of these sequences has limit 0, "the Squeeze Theorem" yields the result.

As for your second question, keep in mind that $\arctan(n)$ is the angle strictly between $-\pi /2$ and $\pi /2$ whose tangent is $n.$ If the tangent of an angle is getting very positive, which angle is that angle approaching. The answer is obvious if you look at a table of values for the tangent (an advantage those of us of a certain generation have over the "calculator generation").

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  • $\begingroup$ Thank you for your solutions. However, please note that the 1 + sin^2 (n) /n is not a sequence. But it is a series. So, the Squeeze Theorem won't work here. Right? $\endgroup$ – Big Boy May 6 '17 at 20:08
  • $\begingroup$ @BigBoy - In your question, you specifically asked what to do with $(1+ \sin ^2(n))/n$ as a sequence $\endgroup$ – Chris Leary May 7 '17 at 0:38
  • $\begingroup$ I stated that he sum of (1 +sin^2(n))/n from n =1 to n=00 and so it is a series not a sequence. Note the word SUM. $\endgroup$ – Big Boy May 7 '17 at 2:01
  • $\begingroup$ @BigBoy - Sorry. I was still concentrating on the $\arctan$ question. The series diverges by comparison with $\sum 1/n.$ You cannot use the divergence test because the terms of the series go to 0. I assume when you sum from $n=1$ to $n=00$ you mean $n=0$ to $n=\infty$? $\endgroup$ – Chris Leary May 7 '17 at 3:47
  • $\begingroup$ I do not think that the comparison test will work though. Since the sum of (1+ sin^2(n))/n is less than or equal to 2/n and 2/n diverges, we can't draw any conclusion about the one we are looking for. So the C.T. is inconclusive. So, the question is still open. I tried the limit comparison test with no luck and the Integral test is not easy and probably not applicable.. $\endgroup$ – Big Boy May 7 '17 at 16:59
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Terminology: The sequence $\{1/n\}_n=(1,1/2,1/3,...)$ converges to $0.$ The series $\sum_n1/n$ diverges. That is, the sequence $(s_n)_n ,$ where $s_n=\sum_{j=1}^n1/j,$ does not converge. $$1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+...>$$ $$>1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+...=1+1/2+1/2+1/2+...\to\infty.$$

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