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P-series math problem

  • Thread starter Benny
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  • #1
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Just a quick question on this series.

[tex]
\sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}}
[/tex]

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 ( [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}} [/tex]) ) 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, [tex]\frac{1}{{\sqrt {n + 1} }} < \frac{1}{{\sqrt n }}[/tex] 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?
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
3,021
2
Benny said:
Just a quick question on this series.

[tex]
\sum\limits_{n = 1}^\infty {\frac{1}{{n^p }}}
[/tex]

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 ( [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}} [/tex]) ) 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, [tex]\frac{1}{{\sqrt {n + 1} }} < \frac{1}{{\sqrt n }}[/tex] 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

[tex]\sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n + 1} }}} [/tex]

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.
 
  • #3
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hello there

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

[tex]\sum\limits_{n = 1}^\infty {\frac{1}{n^s} [/tex]

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
 
  • #4
584
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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.
 
  • #5
matt grime
Science Advisor
Homework Helper
9,395
3
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.
 

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