I'm answering question 19 of chapter two of Spivak's Calculus and I can't seem to think of a way of doing it. I don't want to look up the answer so I thought I'd ask for a hint as to the general direction I should go in. The question states:

If $h>-1$, then $(1+h)^n\geq1+nh$, why is this trivial for $h>0$?

I think I have proved it for $h>0$ using the binomial expansion which eventually leads to:

$\sum\limits_{i=2}^n { \binom{n}{k}h^k}\geq0$ which I believe is true given $h>0\implies h^k>0$ and $\binom{n}{k} \gt0$ for all $k$ in the range $2\leq k \leq n$.

I can't see how this can generalise easily to the case with $-1<h\lt0$ since obviously I can't use the same logic above. I thought about induction but concluded that it wouldn't work since the range only has non-integer values of $k$. It seems intuitively true since $\left|{h^k}\right|>\left|{h^{k+1}}\right|$ and therefore the terms get smaller and smaller but this doesn't take into account the binomial coefficients. So hopefully this isn't a duplicate, I didn't want to browse too much since I don't want the answer just a hint!

Thanks in advance.


We proceed by induction on $n$. The $n=1$ case is trivial: it is just $1+h\ge 1+h$. Suppose the result holds for $n$. Then $$\begin{align} (1+h)^{n+1} &=(1+h)(1+h)^n\\ &\ge (1+h)(1+nh)\\ &=1+(n+1)h+nh^2\\ &\ge 1+(n+1)h \end{align}$$ as desired. Thus by induction this holds for all $n$.

Note: The hypothesis $h\ge -1$ is necessary for the inequality $$(1+h)(1+h)^n\ge (1+h)(1+nh)$$ since for this to hold, we need $1+h>0$.

  • $\begingroup$ Thanks! I now realise than in the question I said that I couldn't use induction because $h$ wasn't an integer but of course that's not the variable that's used in the induction,like you've shown, it's $n$. The question now makes sense with regards to stipulating $h>-1$. Again Thanks for the help. $\endgroup$
    – Jay
    Dec 22 '13 at 16:32
  • $\begingroup$ @Jay No problem. $\endgroup$ Dec 22 '13 at 16:47

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