Divergence Test for an Infinite Series (General question)

[MHCFesh]

This might seem like a rudimentary question but when trying to prove divergence (or even convergence) of an infinite series does the series always have to start at n = 1?

For example would doing a test for [itex]\sum^{∞}_{n=1}\frac{1}{n}[/itex] be any different from [itex]\sum^{∞}_{n=0}\frac{1}{n}[/itex]

 

[micromass]

This might seem like a rudimentary question but when trying to prove divergence (or even convergence) of an infinite series does the series always have to start at n = 1?

No

For example would doing a test for [itex]\sum^{∞}_{n=1}\frac{1}{n}[/itex] be any different from [itex]\sum^{∞}_{n=0}\frac{1}{n}[/itex]

Well, in this special case,

[tex]\sum_{n=0}^{+\infty} \frac{1}{n}[/tex]

is ill-defined at ##n=0## (division by zero), so we need to exclude it. So you can't start at ##0## here. But doing a test at

[tex]\sum_{n=1}^{+\infty}\frac{1}{n}[/tex]

or at

[tex]\sum_{n=10000}^{+\infty}\frac{1}{n}[/tex]

is the same thing. It won't affect convergence.

 

[MHCFesh]

OH, that was silly of me. But thanks a lot for taking your time to explain!

 

[Mark44]

And more generally, what happens in the first few (finite number of) terms at the beginning of the series doesn't affect the convergence or divergence of the series. So these two series have the same behavior:
$$ \sum_{n = 1}^{\infty}a_n$$
and
$$ \sum_{n = k}^{\infty}a_n$$
In the latter series, k is assumed to be larger than 1.