Is the following series convergent?

Question

Answer True/False to the statements.

I found that the limit is 0, and have tried ratio test (L=1) and comparison test, but am unsure how to prove it converges/diverges.

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Answer 1 · 2018-06-13T11:02:13

The series:

#sum_(n->oo) 1/(n+sinn)#

is divergent.

As:

#-1 <= sinx <= 1#

we have that:

#1/(n+1) <= 1/(n+sinn) <= 1/(n-1)#

Based on the squeeze theorem we can therefore immediately determine that:

#lim_(n->oo) 1/(n+sinn) = 0#

Consider now the series:

#sum_(n->oo) 1/(n+1)#

As:

#lim_(n->oo) (1/n)/(1/(n+1)) = lim_(n->oo)(n+1)/n = 1#

based in the limit comparison test it must have the same character as the harmonic series that we know to be divergent.

But as:

#1/(n+1) <= 1/(n+sinn)#

then also the series:

#sum_(n->oo) 1/(n+sinn)#

is divergent by direct comparison.