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Reply 1

You can show it's cauchy by noting

|x_n - x_m| =< |x_n - x_(n-1)| + ... + |x_(m+1)-x_m|

< 1/2^(n-1) + ... + 1/2^m

< 1/2^(m-1) -> 0 as m,n -> &#8734;

Reply 2

But why is that :
1/2^(n-1) + ... + 1/2^m < 1/2^(m-1)

Reply 3

karola
But why is that :
1/2^(n-1) + ... + 1/2^m < 1/2^(m-1)
This is an old thread! Anyway, I'm not sure it was what the other poster meant, but you can sum the series on the LHS as a G.P.

Reply 4

karola
But why is that :
1/2^(n-1) + ... + 1/2^m < 1/2^(m-1)


Hes missed out loads of steps, check your notes from fridays lecture on Analysis (I assume by the bump of this thread your from UCL :p:)

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