## Does this series converge or diverge ?

I have tried all the tests I know, non-null, comparison, ratio, alternating series test. None have conclusive results. Any ideas on how to proceed.
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 When you have a $-1^{n}$ in an infinite series like this, you are going to have to use the alternate series test. 1) Prove that the limit to infinity of $\frac{1}{\sqrt{N+3}}$ goes to zero. 2) Show that $\frac{1}{\sqrt{N+3}}$ is a decreasing sequence. If you can show those two things, then you know that the series is convergent :D

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 Quote by jsewell94 When you have a $-1^{n}$ in an infinite series like this, you are going to have to use the alternate series test. 1) Prove that the limit to infinity of $\frac{1}{\sqrt{N+3}}$ goes to zero. 2) Show that $\frac{1}{\sqrt{N+3}}$ is a decreasing sequence. If you can show those two things, then you know that the series is convergent :D
No, having $-1^n$ would give no useful information. However, having $(-1)^n$ would be helpful.

RGV

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## Does this series converge or diverge ?

Take the modulus of the entire expression, and then you'll end up with $\frac{1}{\sqrt{n+3}}$. Since the latter converges to zero, therefore, so does the original expression. The series converges.

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 Quote by sharks Take the modulus of the entire expression, and then you'll end up with $\frac{1}{\sqrt{n+3}}$. Since the latter converges to zero, therefore, so does the original expression. The series converges.
No. The sequence $\frac{1}{\sqrt{n+3}}$ converges to zero. The series is the sum of all those terms. It diverges.
 It diverges conditionally.
 It diverges absolutely but converges conditionally.
 Recognitions: Gold Member OK, to make it less confusing, here's a suggested solution... Using the comparison test: $$u_n=\frac{1}{\sqrt{n+3}}$$ For n large, $$v_n=\frac{1}{\sqrt n}$$ Using the p-series test, $v_n$ diverges. Therefore, since $u_n$ is less than $v_n$, $u_n$ also diverges. Hence, its sequence converges but its series diverges. Is this correct?

 Quote by sharks OK, to make it less confusing, here's a suggested solution... Using the comparison test: $$u_n=\frac{1}{\sqrt{n+3}}$$ For n large, $$v_n=\frac{1}{\sqrt n}$$ Using the p-series test, $v_n$ diverges. Therefore, since $u_n$ is less than $v_n$, $u_n$ also diverges. Hence, its sequence converges but its series diverges. Is this correct?
This should not be so complicated, its a past exam question in the "easier" section worth 3 marks in a 2 hour exam.

I was thinking about using the fact that "An absolutely convergent Series Converges".

Hence I used the integral test on $$u_n=\frac{1}{\sqrt{n+3}}$$ which equals infinity meaning it diverges.

So that was a fail !

Now the only other test is the alternating series test, clearly 1/sqrt(x) is decresing as x->inf, and the limit goes to zero.

So by the alternating series test the series converges ?(Basially what the second guy said)

Any logical inconsistencies with that proof ?

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 Quote by sid9221 This should not be so complicated, its a past exam question in the "easier" section worth 3 marks in a 2 hour exam. I was thinking about using the fact that "An absolutely convergent Series Converges". Hence I used the integral test on $$u_n=\frac{1}{\sqrt{n+3}}$$ which equals infinity meaning it diverges. So that was a fail ! Now the only other test is the alternating series test, clearly 1/sqrt(x) is decresing as x->inf, and the limit goes to zero. So by the alternating series test the series converges ?(Basially what the second guy said) Any logical inconsistencies with that proof ?
That's exactly what you want for a proof.

 Quote by Dickfore It diverges conditionally.
Sorry, it converges conditionally.

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 Quote by Bohrok It diverges absolutely but converges conditionally.
Now, i'm confused. I don't think i completely understand the meaning of those words.
 A series is said to coverge/diverge absolutely iff the series of absolute values: $$\sum_{n = 1}^{\infty}{\vert a_n \vert}$$ converges/diverges. There is a Theorem that states that if a series converges absolutely, then it also converges ordinarily. The converse, however, is not true. Some alternating series satisfy the criterion that: $$\sum_{n = 1}^{\infty}{a_n}$$ converges, but the series diverges absolutely. For these series, it is said that they converge conditionally.
 Recognitions: Gold Member Science Advisor Staff Emeritus Actually, I don't believe I have ever seen the phrase "diverge conditionally". A series $\sum a_n$ converges conditionally if it converges but $\sum |a_n|$ does not converge. sid9221, you say in your first post that you tried the "alternating series" test and it did not work. Could you explain what you did? It looks to me like it works nicely.
 Recognitions: Gold Member Too much confusion. I think i'll just stick with what i know. No offence.
 Mentor You can also combine pairs of terms: n = 1 & 2, n = 3 & 4, n = 5 & 6, etc. $\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n}{\sqrt{n+3\,}} =\sum_{n = 2,4,\dots}^{\infty}\left(\frac{(-1)^n}{\sqrt{n+3\,}}+\frac{(-1)^{n-1}}{\sqrt{n-1+3\,}}\right)$$\displaystyle =\sum_{k = 1}^{\infty} \left(\frac{1}{\sqrt{2k+3\,}}+\frac{-1}{\sqrt{2k+2\,}}\right)$Combine the rational expressions into one. Use comparison test & p-test.

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