1
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Determine the series:

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

This wouldn't be a alternating series since there is no $(-1)^n$. And I don't think taking a risk of using The Ratio Test would be worth the gamble. I tried taking the limit, in order to use the limit comparison test, but I no longer think it is possible since there are factorials.

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  • $\begingroup$ First rewrite the terms as $\frac{1}{\sqrt{(n+2)(n+1)}}$, and then use the Comparison Test. $\endgroup$ – user84413 Nov 19 '14 at 1:04
  • $\begingroup$ Really? Alright, is that usually helpful with factorial series in general or just this one in particular? $\endgroup$ – Overclock Nov 19 '14 at 1:04
  • $\begingroup$ Ok, so if I was to re-write it for the comparison test then that would be smaller than my original series, so I would be hoping for that to diverge? $\endgroup$ – Overclock Nov 19 '14 at 1:06
  • $\begingroup$ Don't use the Ratio Test, it will be inconclusive. The thing behaves more or less like $\frac{1}{n}$ (limit comparison) or alternately the $n$-th term is bigger than $\frac{1}{n+2}$, so we have divergence. $\endgroup$ – André Nicolas Nov 19 '14 at 1:08
  • $\begingroup$ Yes I got it equal to 1 with the ratio test, unfortunately. $\endgroup$ – Overclock Nov 19 '14 at 1:09
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Notice that the summand is $$\frac{1}{\sqrt{(n + 1)(n + 2)}}.$$

Then, notice that $$\frac{1}{\sqrt{(n + 1)(n + 2)}} > \frac{1}{\sqrt{(n + 2)(n + 2)}} = \frac{1}{n + 2}.$$

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  • $\begingroup$ And we know that it is also less than the original sequence, meaning the whole thing will converge, towards 0 as n goes to infinity right? $\endgroup$ – Overclock Nov 19 '14 at 1:11
  • $\begingroup$ Well, the sequence of terms converges to $0$. But the original series is $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{(n + 1) (n + 2)}}$, which is bounded below by the (divergent) series $\sum_{n = 1}^{\infty} \frac{1}{n + 2}$, and so the original series diverges. $\endgroup$ – Travis Nov 19 '14 at 1:21
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Hint: $\dfrac{1}{\sqrt{(n+1)(n+2)}} \geq \dfrac{1}{n+2}$

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  • $\begingroup$ How do you reach this generalization ? How is this process called ? $\endgroup$ – user2485710 Nov 19 '14 at 3:08
  • $\begingroup$ @user2485710 $$\frac1{\sqrt{(n+1)(n+2)}}\ge\frac1{\sqrt{(n+2)(n+2)}}=\frac1{n+2}$$ The $\ge$ is justified because we are increasing the denominator, which can only make it smaller $\endgroup$ – Justin Nov 19 '14 at 3:50
  • $\begingroup$ @Quincunx I was assuming some kind of technique for doing approximations; I'm noticing that it's nothing like it. Thanks anyway . $\endgroup$ – user2485710 Nov 19 '14 at 14:15

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