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[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?