Proving Convergence of $(-1)^n/\sqrt{n+2}$ Series

[frasifrasi]

(-1)^n/sqrt(n+2) , How do I show that it is conditionally convergent?

The absolute value of the series, 1/n^(1/2) diverges, so shouldn't the series diverge?


--Also, I have never encountered a problem like this, so here is another question:

n(1+n^2)^p , for which values of p does the series converge?
I tried the integral telst by letting u = 1 + n^2, but could not get the answer...

 

[HallsofIvy]

(-1)^n/sqrt(n+2) , How do I show that it is conditionally convergent?

The absolute value of the series, 1/n^(1/2) diverges, so shouldn't the series diverge?

Do you understand what "conditionally convergent" means? By showing that The series with absolute values diverges you have shown that the series is NOT "absolutely convergent". But a series with alternating sign is a whole different matter: use the "alternating series test": if a series is of the form (-1)ⁿa_n and {a_n} is a decreasing sequence, then the series converges. A series that converges but does NOT converge absolutely is called "conditionally convergent".

--Also, I have never encountered a problem like this, so here is another question:

n(1+n^2)^p , for which values of p does the series converge?
I tried the integral telst by letting u = 1 + n^2, but could not get the answer...

??Do you mean a fraction with that in the denominator? If so then go ahead and multiply it out: you get a polynomial with leading term n^{2p+ 1}. It's not too difficult to show that no matter what the other terms are, eventually that fraction is smaller than 2/n^2p+1 and that converges, absolutely for 2p+1> 1 or p> 0.
If you actually do mean the series n(1+ n²)^p then it obviously does not converge for p> 0. If p is negative, it can be written in the form n/(1+ n²)^-p. Now the denominator will have leading coefficient n^2p and, dividing both numerator and denominator by n, is of the form 1/(n^2p-1+ ...). That is clearly larger than 2/n^2p-1 and so converges absolutely for 2p-1> 1 or p> 2.

 

[frasifrasi]

The answer should be p < -1...
That is the series given (n(1+ n^2)^p ), there is no fractions involved.Thank you for the clarification on the first question, now the only thing I still don't get is how to do the second question. I tried using the ratio test, but it didn't get anywhere.

 

[HallsofIvy]

Sorry, I had my signs wrong! n/(1+n^2)^-p is, when you multiply that out, the same as n/(n^(-2p+ other terms) which is the same as 1/(n^-2p-1+ other terms). You can "compare" that to 1/n^-2p-1 which, by the "p test" converges absolutely for -2p-1> 1 or -2p< 2 or p< -1.

 

[frasifrasi]

Thanks, but when I multiply what out?

 

[HallsofIvy]

$(1+ x^2)^{-p}$, of course! (Remembering that, in this case, -p is positive.)

 

[frasifrasi]

How do I multiply that out? I appreciate your patience but it is just not clear to me...

 

[HallsofIvy]

How would you multiply $(1+ x^2)^2$? $(1+ x^2)$?

Notice that the only thing you need to now its the leading term- the one with highest exponent.

 

[frasifrasi]

for that multiplication, $(1+ x^2)^2$, shouldn't the highest exponent be x^4, why do you get n^(-2p-1) for the series, where does the -1 come from?

Oh, I see you divided it by n, so it fits the form 1/n...

 

[frasifrasi]

You flipped the > sign too early in post 4.