The series \sum_{n=1}^\infty n^kr^n coverges when r

Holly Guerrero 2021-12-14 Answered
The series n=1nkrn coverges when r(0,1) and diverges when r>1. This is true regardless of the value of the constant k. When r=1 the series is a p-series. It converges if k<1 and diverges otherwise Each of the series below can be compared to a series of the form n=1nkrn. For each series determine the best value of r and decide whether the series converges.
I am stuck on this question n=1(3n2+4n+22n7n+2+4n+5n)2
I tried using 22n7n+2 and ended up with a value of r of 147 but this didn't work.
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soanooooo40
Answered 2021-12-15 Author has 35 answers
Hint:
4n2+4n+22n3n2(n+)
7n+2+4n+5n7n+2(n+)
the general term of your series is equivalent to
9n472n+4
and has the same nature as
n4(149)n

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limacarp4
Answered 2021-12-16 Author has 39 answers

Note that 
limn(3n2+4n+22n7n+2+4n+5n)2n472n=limn(7n(3n2+4n+22n)n2(7n+2+4n+5n))2 
=limn(3+4n1+22nn272+4n7n+5n7n)2 
=3274 
So, since the series n=1n472n converges, your series converges too.

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