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How would you prove by induction that all members of this sequence lie in the interva

How would you prove by induction for the sequence defined by a1=2a_{1}=2 and an+1=12(an+2a1)a_{n+1}=\dfrac {1}{2}\left( a_{n}+\dfrac {2}{a_{1}}\right) that an∈[1,2]a_{n}\in[1,2]?

Thanks


Posted from TSR Mobile
(edited 9 years ago)
Original post by Brian Moser
How would you prove by induction for the sequence defined by a1=2a_{1}=2 and an+1=12(an+2a1)a_{n+1}=\dfrac {1}{2}\left( a_{n}+\dfrac {2}{a_{1}}\right) that an∈[1,2]a_{n}\in[1,2]?

Thanks


Posted from TSR Mobile


Standard methodology for induction. For the actual P(n) imples P(n+1) bit:

Build the function up bit by bit.

If ana_n is in [1,2], then 2an\frac{2}{a_n} is in .... etc.

Edit: Just noticed you had an a_1 in the denominator. The principle is the same however.
(edited 9 years ago)

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