I need help on question 5.
We can write r = 1+x. Then we have (1+x)^n >= 1+nx i.e.
r^n >= 1 + rn - n
But how does this help us?
Also, would the following argument work?
r^n = exp(n*log r).
log r is clearly negative because of our value of r. So now it's obvious (and we should be able to prove it using the series expansion of exp) that as n goes to infinity, this goes to zero.
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- Thread Starter
- 27-01-2010 22:27
- 27-01-2010 22:59
- 28-01-2010 02:29
In regard to the post above, I think it's more natural to "do something to r to make it greater than 1", so to speak.
Spoiler:Show1/r > 1, so 1/r = (1+x) for some x > 0. Then use 4b.