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Using principle of mathematical induction, prove that Prove the property coined $\mathcal P (n)$ $$(1 + 1/n)^n \le n+1$$ When $n =1$, $LS =2$, $RS =2$

Hence $\mathcal P (1)$ is true. Let $\mathcal P (k)$ be true Then I have to prove $$(1 + 1/k)^k \le k+1$$

#stuck from here onwards. Cannot figure out a way to bring $k+1$... tried multiplying $k+1$ on both sides and obviously failed

Marine Galantin
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pravakar wang
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    This is a homework type question which posted as such will attract downvotes and will eventually be closed. Please show what you have tried so far and provide some context. – Sam Jun 29 '20 at 15:13
  • Welcome to Stackexchange. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Jun 29 '20 at 15:13
  • If you know you should use induction, you could at least check the base case. – Divide1918 Jun 29 '20 at 15:16
  • Tried it several times. Could not think of a way to prove p(k+1) to be true whenever p(k) is true. – pravakar wang Jun 29 '20 at 15:17
  • Tried multiplying by k+1 on both sides of p(k) – pravakar wang Jun 29 '20 at 15:18
  • P(1) is true, Ofc... but cannot find a way to reach p(k+1) from p(k) – pravakar wang Jun 29 '20 at 15:21
  • @pravakarwang Maybe edit the question and add the base case (${p(1)}$) attempt in the question and explain that you were struggling with ${p(k+1)}$ also in the question? Then the question would be considered to have context – Riemann'sPointyNose Jun 29 '20 at 15:22
  • You should use $\ln(1+x)\leq x$ – EDX Jun 29 '20 at 15:23
  • There we go, nice. You can see my answer below – Riemann'sPointyNose Jun 29 '20 at 15:27

2 Answers2

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So, the base case is indeed true. Now, assume

$${\left(1+\frac{1}{n}\right)^{n}\leq n+1}$$

Then clearly

$${\Rightarrow \left(1+\frac{1}{n+1}\right)^{n}\leq n+1}$$

(because ${\frac{1}{n+1} < \frac{1}{n}}$). Now from our assumption,

$${\left(1+\frac{1}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+1}\right)^{n}\leq \left(1+\frac{1}{n+1}\right)(n+1)=n+2}$$

And so by the principle of Mathematical induction, ${\forall\ n \in \mathbb{N}}$ we have

$${\Rightarrow \left(1+\frac{1}{n}\right)^{n}\leq n+1}$$

Riemann'sPointyNose
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A simple way would be to use $\left(1+\frac{1}{n}\right)^n < e$.

Then case $n \ge 2$:

$e < 3 \le n + 1$

Case $n = 1$:

$2 = 2$

thinkingeye
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