Prove that the series `sum_(n=1)^infty 1/n^p` is convergent if p>1 and divergent if p≤1.

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tiburtius eNotes educator| Certified Educator

Integral criterion: If `a_n=f(n)`, where `f(x)` positive, monotonically decreasing and continuous function for `x geq a geq1`, then series `sum_(n=1)^infty a_n` and integral

`int_a^infty f(x)dx`

converge or diverge simultaneously.

Now we apply integral criterion to our series.

`lim_(m->infty) m^(1-p)/(1-p)-1/(1-p)=`

`lim_(m->infty) (m^(1-p)-1)/(1-p)`

We will break this into 3 cases:

p<1 `=> 1-p>0`

`lim_(m->infty)(m^(1-p)-1)/(1-p)=infty=> sum_(n=1)^infty 1/n^p=infty`

Hence the series diverges.

p>0 `=>1-p<0=>p-1>0`

`lim_(m->infty)(m^(1-p)-1)/(1-p)=1/(m^(p-1)(1-p))-1/(1-p)=0-1/(1-p)<infty=>sum_(n=1)^infty 1/n^p<infty`

Hence the series converges.

p=0 `=> 1-p=0`

`lim_(m-> infty)(m^(1-p)-1)/(1-p)=lim_(m->infty)(m^1 m^p-1)/(1-p)=infty=> sum_(n=1)^infty 1/n^p=infty`

Hence the series diverges.

If we put all three cases together we get that the series converges for `p>1` and diverges for `p leq 1`.