Understanding P-Series Math Problems: Convergence and Divergence Explained

[Benny]

Just a quick question on this series.

<br /> \sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}} <br />

It coverges of p > 1 and diverges for all other values of p.

In one of the examples in my book some things are said and there is a line which says observe that the series ( \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}}) ) is a p-series with p = (1/2) < 1 so it diverges. But there is an extra "1" inside the radical so can it still be regarded as a p-series? Also, \frac{1}{{\sqrt {n + 1} }} &lt; \frac{1}{{\sqrt n }} since n is a natural number. So the comparison test wouldn't really tell us anything about the 1/sqrt(n+1) series would it, because the series with 1/sqrt(n) diverges.

Even so, can it still be concluded that 1/sqrt(n+1) is a p-series with p = (1/2) < 1 and so it diverges? The comparison test wouldn't seem to work and I can't really think of any other series to compare to 1/sqrt(n+1). Well apart from something like 1/((n)^(9/10)) but my book seems to have drawn that the conclusion that the series involving 1/sqrt(n+1) is divergent by comparison with 1/sqrt(n). Is that a valid approach?

 

[OlderDan]

Just a quick question on this series.

<br /> \sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}} <br />

It coverges of p > 1 and diverges for all other values of p.

In one of the examples in my book some things are said and there is a line which says observe that the series ( \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}}) ) is a p-series with p = (1/2) < 1 so it diverges. But there is an extra "1" inside the radical so can it still be regarded as a p-series? Also, \frac{1}{{\sqrt {n + 1} }} &lt; \frac{1}{{\sqrt n }} since n is a natural number. So the comparison test wouldn't really tell us anything about the 1/sqrt(n+1) series would it, because the series with 1/sqrt(n) diverges.

Even so, can it still be concluded that 1/sqrt(n+1) is a p-series with p = (1/2) < 1 and so it diverges? The comparison test wouldn't seem to work and I can't really think of any other series to compare to 1/sqrt(n+1). Well apart from something like 1/((n)^(9/10)) but my book seems to have drawn that the conclusion that the series involving 1/sqrt(n+1) is divergent by comparison with 1/sqrt(n). Is that a valid approach?

Try rewriting

\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}}

in terms of m = n + 1. You may want to have a stand alone term in addition to the resulting infinite series to make a comparison.

 

[steven187]

hello there

the comparision test should tell you enough think of the rieman zeta function

\sum\limits_{n = 1}^\infty {\frac{1}{n^s}

it converges for all s>1 and diverges for all s<=1, by the way try the "limit comparision test" ,you should try and compare it to something that behaves like it such as 1/sqrt(n) which is divergent.

steven

 

[Benny]

Yeah, that might have been the method that my book used. It makes sense to use the limit comparison test for this particular series. Thanks for the help.

 

[matt grime]

why comparison? nothing to do with comparison, it's far simpler than that

one sum is

1 + (1/2)^{1/2} + (1/3)^{1/2}+(1/4)^{1/2}+...

the other is

(1/2)^{1/2} + (1/3)^{1/2}+(1/4)^{1/2}+...

so obvlously on diverges iff the other does.