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I have been working a bit on series and came across two problems I couldn't solve:

Determine if the series diverge or converge conditionally/absolutely:

1) $\displaystyle \sum_{n=1}^ \infty \frac{\sqrt n +1} {(n^2 +n +1) }$, I'm completely clueless on what to do on this one

2) $\displaystyle \sum_{n=1}^ \infty \frac{(1+n!)}{(n+1)!}$

I at first thought maybe the ratio test would be useful since it has factorials, but I got 1 with it, so I don't know if that's the way to go about this problem.

Any tips or solutions? Thanks in advance!

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    $\begingroup$ For the first, use limit comparison with $\sum \frac{1}{n^{3/2}}$. For the second, note that the $n$-th term is $\gt \frac{1}{n+1}$, and use the divergence of the harmonic series. Or use limit comparison with $\sum \frac{1}{n}$. $\endgroup$ – André Nicolas Feb 24 '15 at 5:57
  • $\begingroup$ If limit comparison is not in the toolkit, note that in the first, the top is $\le 2\sqrt{n}$ and the bottom is $\gt n^2$, so the $n$-th term is less than $\frac{2}{n^{3/2}}$. $\endgroup$ – André Nicolas Feb 24 '15 at 6:04
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For 1) $\displaystyle \frac{\sqrt{n}+1}{n^2+n+1}\leq \frac{\sqrt{n}+\sqrt{n}}{n^2}=\frac{2}{n^\frac{3}{2}}$. So converges by comparison test.

For 2), $\displaystyle \frac{1+n!}{(n+1)!}=\frac{1}{(n+1)!}+\frac{1}{n+1}\geq \frac{1}{n+1}$. So diverges by comparison test.

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  • $\begingroup$ $\displaystyle \frac{1+n!}{(n+1)!}=\frac{1}{(n+1)!}+\frac{n!}{(n+1)!}=\frac{1}{(n+1)!}+\frac{n!}{(n+1)n!}$ $\endgroup$ – Extremal Feb 24 '15 at 6:16
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For problems like part 1, you can 'simplify' the fraction by considering only the biggest terms in the numerator and denominator. So when you see:

$$\frac{\sqrt{n} + 1}{n^2 + n + 1}$$

You should imagine:

$$\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$$

Which converges using the p test.

To make this approach rigorous, let $a_n = \frac{\sqrt{n} + 1}{n^2 + n + 1}$ and $b_n = n^{-3/2}$, and compute:

$$\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \lim_{n \rightarrow \infty}\frac{\sqrt{n} + 1}{n^2 + n + 1}\cdot n^{3/2} = \lim_{n \rightarrow \infty} \frac{n^2 + n^{3/2}}{n^2 + n + 1} = 1$$

Hence the series $\sum a_n$ converges if and only if $\sum b_n$ converges (this method works whenever $a_n$ and $b_n$ are positive).

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