Question

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n+1} }{ n+5 }\] Apparently this series converges, but is it absolutely or conditionally?

Answer 1

Decreasing toward zero and alternating. -- At least Conditionally.

Integral test: \(\sum\limits_{n=1}^{s-1}\dfrac{1}{n+5} > \int\limits_{1}^{s}\dfrac{1}{x+5}\;dx\)

Answer 2

you must take the ratio test which is:

\[\lim_{n->\infty} \left| \frac{ a_{n+1} }{ a_{n} } \right|\]

if

\[\lim_{n->\infty} \left| \frac{ a_{n+1} }{ a_{n} } \right|\] diverges but

\[\lim_{n->\infty} \frac{ a_{n+1} }{ a_{n} } \] converges

it is CONDIITIONALLY convergent.

if \[\lim_{n->\infty} \left| \frac{ a_{n+1} }{ a_{n} } \right|\]

converges, it implies that both converge (since the criteria is limit < 1 and the only case it can be less than one is if the non-absolute limit isn't already positive.

if you have follow-up questions let me know :)

Answer 3

oh, very cool

Answer 4

if absolute value converges. automatic absolutely convergent

Answer 5

ok

Answer 6

I can't give 2 best answers?
because you both were really helpfull?

Answer 7

hehe :P

Answer 8

The Ratio Test gives "1", and is, therefore, inconclusive.

Answer 9

true. this test is inconclusive.
however he was apparently told that the series converges (possibly in question), making the ratio test probably sufficient

Answer 10

but that is true and very important. if limit of absolute = 1, test is inconclusive (the case)

Answer 11

A good argument, but "Apparently this series converges" hardly provides this information. Okay, @killerbee, what is the real and complete problem statement?

Answer 12

They just said test the series and say why it converges absolutly or conditinally

Answer 13

so it dosn't converge conditinally?

Answer 14

And see "if"?

Read my first post. It provides all necessary information to determine Conditional Convergence.

Answer 15

it does. they probably expected you to use the ratio test.

however tk isn't wrong that it isn't the optimal approach

Answer 16

Converging Absolutely does not mean it doesn't converge conditionally. This is why I said "at least" conditionally. We may promote it to "absolutely", but we didn't, because we proved otherwise with the integral test (or the ratio test if we can simply assume it converges.)

Answer 17

oh

Answer 18

so because the ratio test proves I, that's why it converges conditinally?

Answer 19

No. A ratio test of "1" provides no information.

The integral test showed that the absolute sum fails to converge, therefore it cannot be absolutely convergent.

Absolute convergence is an ADDITIONAL requirement over conditional convergence, not a whole new set of requirements.

Answer 20

ok, you've been so helpfull. I'm going to start another post so I can you a metal :) TY　ｓｏ　ｍｕｃｈ