# Series tests

## Main Question or Discussion Point

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?

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.

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?

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.

Are you referring to tests which tell you whether or not some given infinte 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.

Last edited:
Are you referring to tests which tell you whether or not some given infinte 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) with out affecting whether or not the series (or sequence) converges.
thanks so much thats all I needed to know. An exception would be for example if Its a geometric series and I need to get a number right?

thanks so much thats 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.

Last edited: