Limit evaluation using algebra of sequences and sequence theorems

By making use of only the theorems on sequences (ex: algebra of sequences/cauchy's first theorem of sequences/limit of geometric mean of a sequence etc), how to prove the following: $$lim_{n\to\infty}(1+\frac{1}{n})^n = e$$

• This is a duplicate, but since I'm at my phone I can't look it up. Also does it question mean that the limit exists or that the limit exists and is e for some definition of e? Commented Feb 26, 2019 at 15:45
• The question requires to show that the value of the limit is 'e'. Commented Feb 26, 2019 at 18:25
• What is the definition of 'e'? Commented Feb 26, 2019 at 21:31

$$\left( 1 + \frac{1}{n} \right)^n = e^{n \ln \left(1 + \frac{1}{n} \right)}$$

Now $$n \ln \left(1 + \frac{1}{n} \right) \sim n \times \frac{1}{n} \rightarrow 1$$

• Can you elaborate the second expression where the limit of logarithm expression reduced to 1/n? Commented Feb 26, 2019 at 18:54
• The symbol $\sim$ means "equivalent". Do you know what it is ? Commented Feb 26, 2019 at 19:01
• If you don't know, here is another way to write it : consider the function $f(x)=\ln(1+x)$. You can see that $\frac{\ln(1+x)}{x} =\frac{\ln(1+x)-\ln(1)}{x-0}$, so $\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x}= f'(0)= \frac{1}{1+0} = 1$. In particular, when $n \rightarrow +\infty$, $\frac{1}{n} \rightarrow 0$, so $\lim_{n \rightarrow +\infty} \frac{\ln(1+1/n)}{1/n} = 1$. In other terms, the limit of $n \ln(1 + 1/n)$ is indeed $1$. Commented Feb 26, 2019 at 19:05
• Thank you for the reply and it's clear! I knew the symbol. But i was not trying not to apply any of the concepts other than those related to sequences and few basics. Commented Feb 26, 2019 at 19:18
• Yes I understand. But since your question involves the number $e$, it's natural to use concepts relative to exponential or logarithms... Commented Feb 26, 2019 at 19:20

Simply expand it using binomial theorem cancel out the n's and put n= infinity now all the terms that will be remaining will contain expression like (1-1/n) and so on the term 1/n becomes zero as 1/infinity is zero. Now the resulting expression is the expansion of 'e' thus you get it as e

• It is clear. But since Maclaurin series is required to conclude the last expression to 'e', can we do it in an alternative way? Commented Feb 26, 2019 at 18:35
• I just suggested a way . There are many alternate ways to prove it. Commented Feb 27, 2019 at 15:26