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Determine whether the series $\sum_{n=0}^{\infty} \frac{3n^2 + 2n + 1}{n^3 + 1}$ with n from 0 to infinity converges or diverges.

So far I thought about dividing the numerator by the denominator, but that got very messy. I thought about comparing that to the series of $\frac{1}{k^3 + 1} but then I got stuck.

Also, a related question. A theorem states that if the limit of a series as n approaches infinity is not equal to zero, the series diverges. However it states that the series is from n=1 to infinity. Would it also apply in this case where it goes from n=0 to infinity?

Thanks!

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  • $\begingroup$ I think you mis-read the theorem. It should have stated that the limit of the SUMMAND is non-zero, then the series diverges. When you say 'the limit of a series', that is referring to the limit of the sum of the entire series. And yes it applies for $n=0$ to $infty$ too, provided that the summand is defined at $n=0$. $\endgroup$ – Trogdor Apr 11 '14 at 19:17
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The critical question is whether or not $\sum 3n^2/(n^3+1)$ converges. The other terms, $\sum 2n/(n^3+1)$ and $\sum 1/(n^3 + 1)$ will definitely converge, so the real question is just that first term.

I will claim that the first term diverges because for each term where n > 0, $\sum 3n^2/(n^3+1)$ is greater than the corresponding term of $\sum 3n^2/(3n^3)$ which matches each term of $\sum 1/n$.

Compare with harmonic series...

For the second part of your question, as Trogdor mentioned, yes the same theorem applies, but it is true not for the sequence approaching a value, but the sequence of terms (or summands as Trogdor suggests).

Trivial example: $\sum 1$ The sequence of terms clearly approaches the value 1, which is not 0 (in fact, it actually equals 1 for each term). As you can probably tell, this sequence diverges.

In general, if $\sum_{n=1}^\infty \Phi (n)$ diverges and $\Phi(0)$ is defined, then clearly $\Phi (0) + \sum_{n=1}^\infty \Phi(n)$ diverges, but this last result is just $\sum_{n=0}^\infty \Phi(n)$, so the theorem applies here as well.

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  • $\begingroup$ To me, it is not automatically obvious that ∑3n^2/(n^3+1) is greater than the corresponding term of ∑3n^2/(3n^3). How can you quickly find the appropriate series that it bigger/smaller as needed? $\endgroup$ – Chrysanthemum Apr 11 '14 at 19:46
  • $\begingroup$ Because $n^3+1 < 3n^3$ whenever n > 0. Increasing denominator for each term implies that result decreases. To quickly find appropriate series, the best place to start is to look at the most significant terms of both the numerator and the denominator. $\endgroup$ – Bill Province Apr 11 '14 at 19:58
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The $n$-th term is $\ge \frac{3n^2}{2n^3}$, which is $\frac{3}{2}\cdot\frac{1}{n}$. Now compare with the harmonic series to deduce divergence.

As to your second question, the convergence/divergence of a series is not affected if we alter a finite number of terms. So whether we sum from $0$ to $\infty$ or from $1$ to $\infty$ makes no difference.

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