# Convergence and divergence of series?

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#1
Solved.
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2 years ago
#2
first one converge I guess,

because the denominator each term of the first series is atleast epsilon greater than the respective term' s denominator of the harmonic series, then by the harmonic series test, it should converge I believe. This is when b is assumed to be greater than 0. But when b is less than 0, the series sum diverges for the reason that its convergence is 'worse' than that of the harmonic series. At b=0, the series is harmonic series minus 1 which is infinity too.

for the second one, I guess it will converge too because atleast the term at r at infinity is approaching 0. Now all one needs to check is if the sum of these terms approaching 0 as r tends to infinity converges too. Then you can use binomial expansion on (1+1/r)^(r^2), then multiply that with 1/3^r, then by rearranging this completely positive termed series, we get back terms that look like terms of the harmonic series multiplied by something else (since you have 'b' all positive as r^2 is greater 0), the series will converge.
0
2 years ago
#3
(Original post by sajeth)
first one converge I guess,

because the denominator each term of the first series is atleast epsilon greater than the respective term' s denominator of the harmonic series, then by the harmonic series test, it should converge I believe. This is when b is assumed to be greater than 0. But when b is less than 0, the series sum diverges for the reason that its convergence is 'worse' than that of the harmonic series. At b=0, the series is harmonic series minus 1 which is infinity too.
No. The sum when b = 1 is a well known divergent series.

for the second one, I guess it will converge too because atleast the term at r at infinity is approaching 0. Now all one needs to check is if the sum of these terms approaching 0 as r tends to infinity converges too. Then you can use binomial expansion on (1+1/r)^(r^2), then multiply that with 1/3^r, then by rearranging this completely positive termed series, we get back terms that look like terms of the harmonic series multiplied by something else (since you have 'b' all positive as r^2 is greater 0), the series will converge.
I can't follow what you're trying to say here at all. The (1+1/r)^(r^2) term will behave like e^r, so the fact that 3 > e is critical here. Your argument doesn't seem to acknowledge this at all.
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