Question

Prove that the following series is convergent:
\[\sum_{n=1}^{\infty}{\frac{n}{\sqrt{n^4+n+1}+1}}\]
(Possibly by Limit Comparison Test?)

Answer 1

I don't know if this will help or not, but you can re-arrange this to:\[\frac{n}{\sqrt{n^4+n+1}+1}=\frac{n(\sqrt{n^4+n+1}-1)}{(\sqrt{n^4+n+1}+1)(\sqrt{n^4+n+1}-1)}\]\[=\frac{n(\sqrt{n^4+n+1}-1)}{n^4+n+1-1}\]\[=\frac{n(\sqrt{n^4+n+1}-1)}{n^4+n}\]\[=\frac{\sqrt{n^4+n+1}-1}{n^3+1}\]

Answer 2

alternatively, although this may no be a proof in the strictest sense, you could argue that as 'n' increases, then:\[\sqrt{n^4+n+1}+1\rightarrow \sqrt{n^4}+1\rightarrow n^2+1\rightarrow n^2\]so then we can say that the terms tend to:\[\frac{n}{n^2}=\frac{1}{n}\]which is convergent

Answer 3

The only problem with that last bit is that the sum of 1/n is the harmonic series which would be divergent. Thanks for the re-arrangement though, I'll see if it proves it. Thankyou.

Answer 4

glad to be of "some" assistance - let me know what the solution is please - if you do find one that is :-)

Answer 5

I'd be interested to know these should be done "properly"

Answer 6

I'm such a moron. I'm pretty sure the series diverges. I think I have a proof, if I get it I'll write it out.

Answer 7

thanks - and no, you are not a moron - we all have "moment" of madness :-)

Answer 8

Right.
\[\text{Let }a_n = \frac{n}{\sqrt{n^4+n+1}+1} = \frac{\sqrt{n^4+n+1}-1}{n^3+1}\text{ (Thankyou to asnaseer)}\]\[\text{Let }b_n = \frac{1}{n+1}\]\[\frac{a_n}{b_n} = \frac{\sqrt{n^4+n+1}-1}{n^2-n+1}\]\[\lim_{n\to\infty}{\frac{a_n}{b_n}} = \lim_{n\to\infty}{\left(\frac{\sqrt{n^4+n+1}}{n^2-n+1} - \frac{1}{n^2-n+1}\right)} = 1 - 0 = 1\]\[\text{Since }b_n \text{ is divergent and }\lim_{n\to\infty}{\frac{a_n}{b_n}} = 1\text{, } a_n \text{ is divergent by the Limit Comparison Test}\]
I reaaaally hope that's not wrong.

Answer 9

how do you deduce the limit of:\[\frac{\sqrt{n^4+n+1}}{n^2-n+1}\]as n tends to infinity is 1?

Answer 10

oh wait - I see it now

Answer 11

Slight correction, it should be:
\[\text{Since }\sum_{n=1}^{\infty}{b_n}\text{ is divergent and }\lim_{n\to\infty}{\frac{a_n}{b_n}} = 1\text{, }\sum_{n=1}^{\infty}{a_n}\text{ is divergent by the LCT}\]
Since the fraction contains only powers of n, the limit is the ratio of the highest power on the numerator to the highest power on the denominator. Which is:\[\frac{\sqrt{n^4}}{n^2} = 1\]

Answer 12

yup - got it - and I also now undertand a bit about LCT as I started reading up on it in the meantime. thanks for the explanation.