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Determine wether these series are convergent:

1) $$\sum_{l=5}^\infty \left(\frac{1}{l^2} + \frac {2}{l^3}\right)$$
$$ \left(\frac{1}{l^2} + \frac {2}{l^3}\right)=\frac {l^3+l^2}{l^5} > \frac {1}{l^2}$$
Does this way works for the comparison test?
I don't think because the minorant is convergent, am I right?
Could I do it by this way:
$$ a_l:=\frac{1}{l^2} \text{ and } b_l:=\frac {2}{l^3} $$
and then say that both series $a_l$ and $b_l$ are convergent $p$-series so that the sum also have to be convergent, is that correct or is this only allowed for multiplication with series?

2) $$\sum_{n=1}^\infty \left(\frac {1}{n^2+n+1}\right)$$
With the CT:
$$\left\lvert\frac {1}{n^2+n+1}\right\rvert < \frac{1}{n^2}$$
Because of the harmonic $p$-seriew:
$$\sum_{n=1}^\infty \frac{1}{n^2}$$ which converges,
so 2) is convergent.

3) $$\sum_{m=1}^\infty \left(\frac {4m^3-2m^2-m+5}{m^7+3m^2-2m}\right)$$
Here I have the same problem because I found a series which is majorant which is smaller then the series 3)..
My idea would be to multiply 3) by $1/b_k$ and look at the limit.

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too many questions on same post :((!! – Santosh Linkha Jan 10 '13 at 21:54
perhaps.. I thougth I should write the 3 series because my questions are quite connected for the 3 series. – phil Jan 10 '13 at 21:58
up vote 1 down vote accepted

For 3), you should see that the summand behaves as $4/m^4$ as $m \rightarrow \infty$, so it converges by the comparison test.

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So I can neglect the other terms,in the same way as if I would regard the limit of a sequence? – phil Jan 10 '13 at 21:55
Yes, the other terms are necessarily smaller. – Ron Gordon Jan 10 '13 at 21:59
then if it works for 3), it also works for 1) and i can say that it's convergent because "1)" = $1/l^2$ and that converges? – phil Jan 10 '13 at 22:11
I suppose that is correct. – Ron Gordon Jan 10 '13 at 22:19

$1) \sum \limits_{l=5}^\infty (\frac{1}{l^2} + \frac {2}{l^3})$

This is a sum of two P-series $ - \text{ something }$ with $p>1$, since both converges, it should.

$2) \sum \limits_{n=1}^\infty \frac {1}{n^2+n+1}$
As you reasoned

$3) \sum \limits_{m=1}^\infty (\frac {4m^3-2m^2-m+5}{m^7+3m^2-2m}) < \sum \limits_{m=1}^\infty \frac {4m^3-2m^2-m+5}{m^7} = \sum_{m=1}^\infty \left( \frac{4}{m^4} - \frac{2}{m^5} - \frac{1}{m^6} + \frac{5}{m^7} \right)$

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Thanks now I see it in 3) so to do the comparison test of a fraction with 2 polynoms, I always have to neglect the other terms from lower degree in the denominator and after this I only have regard the different serie terms with the p-test? – phil Jan 10 '13 at 22:31
@phil yep!! that's the idea ,,, whenever they get big $n\to \infty $ the term with the highest power will be dominate others ... so, you can do it. – Santosh Linkha Jan 10 '13 at 22:33

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