Show that the series is convergent

[knv]

1. Show that the series is convergent and then find how many terms we need to add in order to find the sum with an error less than .001

Ʃ (-1)^(n-1)/ √(n+3)

from n = 1---> infinity



2. I took the derivative.



3. f(x) = (x+3)^-1/2
f'(x) = -1/2 (x+3)^-3/2

Then I set up the following

Absolute value (1/(n+1+3)) < .001

n+4 > (1/.001)²

Got 999,997 for the answer. not sure what I am doing wrong. Help!

 

[who_]

There really is no need to be taking derivatives here. You should try to apply the Alternating Series Test. If you don't know what that is, you can find it on Wikipedia. The page has all of your answers.

 

[knv]

would it be 999,998 terms ?

 

[SammyS]

would it be 999,998 terms ?

How did you get that?

The absolute value of the 999,997^th term is indeed 0.001 .

But the series is alternating and the absolute value of subsequent terms if decreasing. If we let S_k represent the k^th partial sum, then I would expect the series to converge to a value very close to midway between S_k and S_k+1, for large k.