Third method: The two sides if the inequality are equal at x=0. The derivative of the RHS is n, and the derivative of the LHS is n(1+x)^(n-1) >= n, since 1+x > 1. So the LHS is increasing at a faster rate than the RHS, so the LHS must always be at least as large as the RHS
Why set x > 0 here rather than > -1 like the principal seems to originally state? is it just a simpler approach and the full proof would have a separate case for between -1 and 0?
Third method:
The two sides if the inequality are equal at x=0. The derivative of the RHS is n, and the derivative of the LHS is n(1+x)^(n-1) >= n, since 1+x > 1. So the LHS is increasing at a faster rate than the RHS, so the LHS must always be at least as large as the RHS
you are not allowed to do that but good try
@@nikos4677why he can’t do that.?
@@jehannabary3872 You are not allowed to differentiate on an inequality.
You could have taken n=0 as your base case
Thanks for the greaet video
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How do you use this to prove a sequence is growing sequence
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Why set x > 0 here rather than > -1 like the principal seems to originally state? is it just a simpler approach and the full proof would have a separate case for between -1 and 0?
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