Question

does this series converge or diverge?
2n+1/n3^n
explain

Answer 1

it converges... using the divergence test:
\[\lim_{n \rightarrow \infty}2n+1/n3^n=\lim_{n \rightarrow \infty}2n/n3^n+1/n3^n=\lim_{n \rightarrow \infty}2/3^n+1/n3^n=0+0\]
=0,
so the series converges

Answer 2

not quite sure if it is right to conclude series converges using divergence test.
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/divergence.html

Answer 3

I suppose anonymous is correct but when the limit is zero using that test it doesn't necessarily mean it converges. Does it? Because the harmonic sequence limit is zero but it's a divergent series.

Answer 4

i was thinking the same.
but you can use comparison test. it's quite simplified above.

Answer 5

@experimentX is very correct. The divergence test is only useful for testing the divergence of a series.

Answer 6

or if you want to be absolutely sure, I will .... so:
2n+1/n3^n=2n/n3^n+1/n3^n=2/3^n+1/n3^n
2/3^n will converge since its a geometric series with ratio 1/3
for 1/n3^n, use ratio test so
\[|(1/(n+1)3^{n+1})(n3^n)|=\lim_{n \rightarrow \infty}|n/3(n+1)|=\lim_{n \rightarrow \infty}|1/3|=1/3\]
so that series is convergent, and since our series is a sum of two convergent series, then our series is also convergent