Vtyjkyuk

I don't want to use the definition, is there an easier way?

Like, in the case of limits of functions, it is pretty easy to say by the composition theorem.

Is there any such theorem in case of sequences too?

Hint:

As exponential function is continuous and "limit of the function"=function of the limit" for a continuous function, so$$lim_n e^1/n=e^lim_n 1/n=e^0=1.$$

$$1<e^frac1n<3^frac1n=(1+2)^frac1n<1+dfrac2nto1$$
as $ntoinfty$.

I figured it out. Since the function whose domain is $mathbbR$ converges to $1$ (By composiotion theorem). The sequence of the same functional rule should, too.

Since, we know that for every $epsilon>0$, there exists $N$, so that $|f(n)-L|<epsilon$ for $n>N$, So for every $epsilon>0$, there exists $N$, so that $|f(n)-L|<epsilon$ for $n in mathbbI>N$. So the sequence converges too.
That is, $$lim_ntoinftya_n=L$$ too.

For $nin Bbb Z^+$ let $e^1/n=1+A_n.$ We have $A_n>0.$ From the Binomial Theorem we have $$e=(1+A_n)^n=sum_j=0^nbinom nj(A_n)^jgeq sum_j=0^1binom nj(A_n)^j=1+nA_n.$$ Therefore $frac e-1ngeq A_n.$ This also works if we replace $e$ with any number greater than $1.$