`a_n = nsin(1/n)` Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

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`a_n=nsin(1/n)`

Apply n'th term test for divergence, which states that, 

If `lim_(n->oo) a_n!=0` , then `sum_(n=1)^ooa_n` diverges

`lim_(n->oo)nsin(1/n)=lim_(n->oo)sin(1/n)/(1/n)`  

Apply L'Hospital's rule,

Test L'Hospital condition:`0/0`

`=lim_(n->oo)(d/(dn)sin(1/n))/(d/(dn)1/n)` 

`=lim_(n->oo)(cos(1/n)(-n^(-2)))/(-n^(-2))`

`=lim_(n->oo)cos(1/n)`

`lim_(n->oo)1/n=0`

`lim_(u->0)cos(u)=1` 

By the limit chain rule,

``

`=1!=0` 

So, by the divergence test criteria series diverges.  

` `

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`a_n=nsin(1/n)`

Apply n'th term test for divergence, which states that, 

If `lim_(n->oo) a_n!=0` , then `sum_(n=1)^ooa_n` diverges

`lim_(n->oo)nsin(1/n)=lim_(n->oo)sin(1/n)/(1/n)`  

Apply L'Hospital's rule,

Test L'Hospital condition:`0/0`

`=lim_(n->oo)(d/(dn)sin(1/n))/(d/(dn)1/n)` 

`=lim_(n->oo)(cos(1/n)(-n^(-2)))/(-n^(-2))`

`=lim_(n->oo)cos(1/n)`

`lim_(n->oo)1/n=0`

`lim_(u->0)cos(u)=1` 

By the limit chain rule,

``

`=1!=0` 

So, by the divergence test criteria series diverges.  

` `

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