Solving Series Tests with n > 1: What to Do?

[icochea1]

I see that most series tests says that it needs to be like

(infinity)
E (An)
n=1

what happens when you get a problem that is for example n=2 or n=3?

 

[sutupidmath]

Well, we start with n=2,3,.. sometimes because of the function at question. But it doesn't make any difference, since the addition or subtraction of a finite number of terms doesn't affect the convergence of a series.

 

[icochea1]

Well, we start with n=2,3,.. sometimes because of the function at question. But it doesn't make any difference, since the addition or subtraction of a finite number of terms doesn't affect the convergence of a series.

so for example if I have to use the Alternating Series test and n=2 I just work the same as if it equaled to 1?

 

[sutupidmath]

sometimes if you are comparing two series, say a, b. it might happen that for example

a<b, only for n> say 4 or sth.

 

[Pere Callahan]

Are you referring to tests which tell you whether or not some given infinite series converges?

If so, the tests apply equally well to sums in which the index of summation starts at some finite integer larger than one because you can always change finitely many terms of a series (or sequence) without affecting whether or not the series (or sequence) converges.

 

[icochea1]

Are you referring to tests which tell you whether or not some given infinite series converges?

If so, the tests apply equally well to sums in which the index of summation starts at some finite integer larger at one because you can always change finitely many terms of a seris (or sequence) without affecting whether or not the series (or sequence) converges.

thanks so much that's all I needed to know. An exception would be for example if Its a geometric series and I need to get a number right?

 

[Pere Callahan]

thanks so much that's all I needed to know. An exception would be for example if Its a geometric series and I need to get a number right?

This is right. For a geometric series you can write

$$\sum_{n=N}^\infty{q^n}=\sum_{n=1}^\infty{q^n}-\sum_{n=1}^{N-1}{q^n}=\frac{q}{1-q}-\frac{q-q^N}{1-q}=\frac{q^N}{1-q}$$

which is (of course) different from what you would expet for a "complete" geometric series, corresponding to N=1 (or N=0, as you like). However, for each natural N, the series converges exactly for all q whose absolute value is strictly less than unity.