How to prove that $$k+1\ge \bigg(1+\frac{1}{k}\bigg)^{k} $$ when $k>2$
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1If $k=-2$, then LHS < RHS. – symmetricuser Oct 07 '14 at 16:55
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1prove when k>2. – Sona Oct 07 '14 at 17:06
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RHS is an increasing sequence bounded up by e – John Oct 07 '14 at 17:06
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I know it, but how to prove without lim, when we had to prove that we hadn't passed bounded sequences and e. – Sona Oct 07 '14 at 17:08
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2Worked it out in my head, but too late in my time zone to post it, but here's a hint: consider the binomial expansion of the RHS. What can you say about the upper bound on each term? How many terms are there? – Deepak Oct 07 '14 at 17:09
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We will prove this with induction.
Consider the base case of $k=2$. Clearly, $3\ge 2.25$.
Now assume that $k+1\ge \left(1+\frac{1}{k}\right)^k$. We want to show that $\left(k+1\right)+1\ge\left(1+\frac{1}{k+1}\right)^{k+1}$. Consider
$$\begin{align} \left(1+\frac{1}{k+1}\right)^{k+1}&=\left(1+\frac{1}{k+1}\right)^k\left(1+\frac{1}{k+1}\right)\\ &\le\left(1+\frac{1}{k}\right)^k\left(1+\frac{1}{k+1}\right)\\ &\le\left(k+1\right)\left(1+\frac{1}{k+1}\right)\\ &=(k+1)+\frac{k+1}{k+1}\\ &=(k+1)+1 \end{align}$$
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