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I am having some trouble understanding how to plug terms into the following sequence. I was given the following,

$$a_1=1, a_{n+1}=\frac{a_n}{\sqrt{n}}$$ for all n greater or equal to 1.

Plugging in terms we get the following,

$$a_1=1$$ $$a_2=a_1=1$$ $$a_3=\frac{a_2}{\sqrt{2}}=\frac{1}{\sqrt{2}}$$ $$a_4=\frac{a_3}{\sqrt{3}}=\frac{1}{\sqrt{6}}$$

How are they getting these terms $a_2,a_3,a_4$?

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Notice that

$$a_2=\frac{a_1}{\sqrt1}=1$$

$$a_3=\frac{a_2}{\sqrt2}=\frac1{\sqrt2}$$

$$a_4=\frac{a_3}{\sqrt3}=\frac1{\sqrt6}$$

Try to prove by induction that

$$a_{n+1}=\frac1{\sqrt{1\cdot2\cdot3\cdots n}}=\frac1{\sqrt{n!}}$$

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  • $\begingroup$ why are the $a_1,a_2,a_3$ terms disappearing in the numerator? $\endgroup$ – jh123 Feb 14 '17 at 15:32
  • $\begingroup$ @jh123 Let's start with $a_2$. Do you understand that $a_2=\frac{a_1}{\sqrt1}$? From there, what is $a_1$? (check definition?) $\endgroup$ – Simply Beautiful Art Feb 14 '17 at 15:33
  • $\begingroup$ why would $a_2$ be equal to that? I thought $a_2=a_1+a_2$? $\endgroup$ – jh123 Feb 14 '17 at 15:34
  • $\begingroup$ Because $a_2$ was defined to be equal to that. Plug $n=1$ into the definition of the sequence. Why would $a_2=a_1+a_2$, which would require $a_1=0$? $\endgroup$ – Ross Millikan Feb 14 '17 at 15:37
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    $\begingroup$ @jh123 But don't forget to take a peak at the second part of my answer:$$a_{n+1}=\frac1{\sqrt{n!}}$$ $\endgroup$ – Simply Beautiful Art Feb 14 '17 at 15:55
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They are using the definition. If you plug in $n=1$ you get $a_2=\frac {a_1}{\sqrt 1}=\frac 11=1.$ Then plug in $n=2$ to get $a_3=\frac {a_2}{\sqrt 2}=\frac 1{\sqrt 2}$ and so on.

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  • $\begingroup$ why are the $a_1,a_2,a_3$ terms disappearing in the numerator? $\endgroup$ – jh123 Feb 14 '17 at 15:33
  • $\begingroup$ Because we plug in the value we know for them based on the earlier calculation. $\endgroup$ – Ross Millikan Feb 14 '17 at 15:36
  • $\begingroup$ so if $a_1=1$ then shouldn't $a_2 = \frac{a_1}{\sqrt{2}}= \frac{1}{\sqrt{2}}$ $\endgroup$ – jh123 Feb 14 '17 at 15:38
  • $\begingroup$ No, because we were taking $n=1$ and the denominator is $\sqrt n$, not $\sqrt {n+1}$ $\endgroup$ – Ross Millikan Feb 14 '17 at 15:56
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We have that for $n \in \mathbb Z^+$, $$a_{n+1} = \frac {a _n}{\sqrt {n}} \tag {1}$$

Thus, using the relation $(1)$ successively, we get, $$a_2 = a_{1+1} = \frac {a_1}{\sqrt {1}} = \frac {1}{1} =1 $$ $$a_3 = a_{2+1} = \frac {a_2}{\sqrt {2}} = \frac {1}{\sqrt {2}} $$ $$a_4 = a_{3+1 } = \frac {a_3}{\sqrt {3}} = \frac {1}{\sqrt {2}\times \sqrt {3}} = \frac {1}{\sqrt {6}} $$

Hope it helps.

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  • $\begingroup$ why is the $a_1$. $a_2$ $a_3$ terms disappearing in the numerator? $\endgroup$ – jh123 Feb 14 '17 at 15:31

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