Hi I have two questions.

First, $\sum_{n=1}^\infty \frac{n}{n^3+1}$.

Is it divergent or convergent? I think it seems like it is positive and decreasing function so we can apply integral test. however, integrating this function is not an easy task.

Is there any other test i can use?

Second, $\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}$.

I can't think of any test,,

Can anyone give me an idea please~

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    $\begingroup$ Don't the terms in the first sum resemble $1/n^2$? And in the second sum, don't they resemble $1/n$? $\endgroup$ – zhw. Apr 17 '15 at 3:34
  • $\begingroup$ So for the first one I can use comparison test so that 1/n^2 is greater than that and is convergent and it must be converegent $\endgroup$ – Nancy Apr 17 '15 at 3:37
  • $\begingroup$ for the second part I think I could use comparison test but can't see which one shall I use $\endgroup$ – Nancy Apr 17 '15 at 3:37

A great starting point on these questions is to "ignore all the small pieces" and see what happens. For instance, for large $n$, the $+1$ in the denominators won't really matter. So the first really looks like $$ \sum_{n \geq 1} \frac{1}{n^2}$$ and the second looks like $$ \sum_{n \geq 1} \frac{1}{n}.$$ Some calculus books have a so-called "limit comparison" test that makes this analysis rigorous. Barring that, let's see if we can do better.

In the first, $$ \sum_{n \geq 1} \frac{n}{n^3 + 1} < \sum_{n \geq 1} \frac{n}{n^3} = \sum_{n \geq 1} \frac{1}{n^2}.$$

In the second, $$\sum_{n \geq 1} \frac{n^2 + 1}{n^3 + 1} > \sum_{n \geq 1} \frac{n^2}{n^3 + n^3} = \frac{1}{2}\sum_{n \geq 1} \frac{1}{n}.$$

  • $\begingroup$ for the second part, how can I show that n^2/n^3+n^3 is less than n^2+1/n^3+1? $\endgroup$ – Nancy Apr 17 '15 at 3:41
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    $\begingroup$ I made the numerator smaller and the denominator bigger. $\endgroup$ – davidlowryduda Apr 17 '15 at 3:42
  • 2
    $\begingroup$ Thanks makes a lot sense to me $\endgroup$ – Nancy Apr 17 '15 at 3:43

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