How to prove $\lim_{n\to\infty} \left(1+\frac{1}{(3n+1)}\right)^n $ = $ e^\frac{1}{3}$

Question

I really just need to know how to prove $ \lim_{n\to\infty} \left(1+\frac{1}{(3n+1)}\right)^n $ = $ e^\frac{1}{3}$

I have tried using the squeeze theorem and tried to find a convergent subsequence.

Any help would be appreciated

Answer 1

Hints:

Either you know the standard limit of $\Bigl(1+\dfrac1n\Bigr)^n$, and you just have to set the substitution $m=3n+1$.

Or you determine the limit of the log.

Answer 2

Hint.

You know that:

$$(1+\frac 1{3n+1})^n=e^{n\log(1+1/(3n+1))}$$

and you can know use the fact:

$$\log(1+x)\underset{x\to 0}{\sim}x$$

to conclude.

More hint : because then $\log(1+1/(3n+1))\underset{n\to \infty}{\sim}1/(3n+1)$.

Answer 3

$$\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 1 }{ 3n+1 } \right) }^{ n } } =\lim _{ n\rightarrow \infty }{ { \left[ { { \left( 1+\frac { 1 }{ 3n+1 } \right) }^{ 3n+1 } } \right] }^{ \frac { n }{ 3n+1 } } } ={ e }^{ \lim _{ n\rightarrow \infty }{ \frac { n }{ n\left( 3+\frac { 1 }{ n } \right) } } }={ e }^{ 1/3 }$$