How do I find the limit of the sequence $v_n=\left(1+\frac{1}{3n+1}\right)^n$? I have to find the limit of
$$v_n=\left(1+\frac{1}{3n+1}\right)^n$$
I think that $v_n$ is a subsequence of $\left(1+\frac{1}{n}\right)^n$. So limit is $e$?
But answer is given as $e^{1/3}$.
Where am I going wrong?
Thanks in advance
 A: Hint:
$$v_n=\left(1+\frac1{3n+1}\right)^{n}=\left[\left(1+\frac1{3n+1}\right)^{3n+1}\right]^{\frac13}\left(1+\frac1{3n+1}\right)^{-\frac13}.$$
A: Let $\gamma = 3n+1$, now, we can write $n$ as $\dfrac{\gamma-1}{3}$. Rewriting $v_n$ we have that $v_{\gamma} = \left(1+\dfrac{1}{\gamma}\right)^{\dfrac{\gamma-1}{3}}$ or $v_{\gamma} = \dfrac{\left(1+\dfrac{1}{\gamma}\right)^{\dfrac{\gamma}{3}}}{\left(1+\dfrac{1}{\gamma}\right)^{\dfrac{1}{3}}}$. Now, taking the limit as $\gamma \rightarrow \infty$ we obtain: $$\dfrac{\lim_{\gamma \rightarrow \infty}\left(1+\dfrac{1}{\gamma}\right)^{\dfrac{\gamma}{3}}}{\lim_{\gamma \rightarrow \infty}\left(1+\dfrac{1}{\gamma}\right)^{\dfrac{1}{3}}}.$$ Which is equal to $$\dfrac{\lim_{\gamma \rightarrow \infty}\left({\left(1+\dfrac{1}{\gamma}\right)^{\gamma}}\right)^{\dfrac{1}{3}}}{1}.$$
Evaluating that limit we obtain that $$\lim_{\gamma \rightarrow \infty}\left({\left(1+\dfrac{1}{\gamma}\right)^{\gamma}}\right)^{\dfrac{1}{3}} = e^{1/3}.$$
A: If you don't see an easy way to immediately convert it to an $e$ answer,  you can always use the fact that logarithms and exponents are continuous so they interchange with limits,  and using the identity $f(x)=e^{\ln (f(x))}$
so
$$\ln (\lim_{n\to \infty}(1 + \frac 1 {3n+1})^n)= \lim_{n\to \infty}\ln ((1 + \frac 1 {3n+1})^n)$$
$$=\lim_{n\to \infty}n\ln ((1 + \frac 1 {3n+1}))$$
Now you have $0\times \infty$ indefinite form, so we convert the times $n$ to divide by $\frac 1 n$
$$=\lim_{n\to \infty}\frac {\ln ((1 + \frac 1 {3n+1}))}{\frac 1 n}$$
Now we have $0/0$,  but L'Hospital's is annoying in this case, so to make it easier we do the variable substitution $x=\frac 1 n$, which convers us to a limit going to 0
$$=\lim_{x\to 0^+}\frac {\ln ((1 + \frac 1 {\frac 3 x+1}))}x$$
Rewriting the fraction inside by multiplying by $\frac x x$ gets us to
$$=\lim_{x\to 0^+}\frac {\ln ((1 + \frac x {3+x}))}x$$
Knowing we are going to do L'Hospitals, to make my derivivative even easier I am going to add and subtract 3 to the numerator of the fraction with x:
$$=\lim_{x\to 0^+}\frac {\ln ((1 + \frac {x+3-3} {3+x}))}x$$
Which lets us split it into $1$ plus the part with the remaining numerator $-3$
$$=\lim_{x\to 0^+}\frac {\ln ((2 + \frac {-3} {3+x}))}x$$
Now finally using L'Hospitals gets us to
$$=\lim_{x\to 0^+}\frac {3(x+3)^{-2}}{((2 + \frac {-3} {3+x}))}$$
Multiplying top and bottom by $(3+x)^2$ gets us to
$$=\lim_{x\to 0^+}\frac {3}{((2(x+3)^2 + (-3)(x+3)))}$$
Evaluating at $x=0$ gets us to $\frac 1 3$
But that wasn't where our limit went to, that's where the natural log of our limit went to.  So to recover the original limit, we take $e$ to that power.
A: $$v_n\,=\,\left(1+\frac{1}{3n+1}\right)^n\,=\,\left(\left(1+\frac{\frac{1}{3}}{n+\frac{1}{3}}\right)^{n+\frac{1}{3}}\right)^\frac{n}{n+\frac{1}{3}}$$
