convergence of this SEQUENCE

[iDoof]

I have the general term of a sequence: {(1 + 3/n)^4n}. Does it converge or diverge? I'm not sure how to go about finding its limit...could someone kindly explain it to me?


thanks!!

 

[galactus]

Does this look familiar?:

\(\displaystyle \lim_{n\to\infty}\(1+\frac{1}{n})^{n}=e\)

You could start by letting, say, $\displaystyle t=\frac{3}{n}$

Then $\displaystyle n=\frac{3}{t}$

\(\displaystyle \lim_{n\to\infty}(1+\frac{3}{n})^{n}=\lim_{n\to\0}(1+t)^{\frac{3}{t}}\)

\(\displaystyle =\lim_{n\to\0}[(1+t)^{\frac{1}{t}}]^{3}=e^{3}\)

$\displaystyle (e^{3})^{4}=e^{12}$

 

[soroban]

Hello, iDoof!

I have the general term of a sequence: .$\displaystyle a_n\;=\;(1\,+\,\frac{3}{n})^{4n}$
Does it converge or diverge?

Are you familiar with the definition of e? . \(\displaystyle \L\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)

We have: .$\displaystyle \left(1\,+\,\frac{3}{n}\right)^{4n}$

. . Rewrite the fraction: .$\displaystyle \frac{3}{n}\:=\:\frac{1}{\frac{n}{3}}$

. . Multiply the exponent by $\displaystyle \frac{3}{3}:\;\;4n\cdot\frac{3}{3}\:=\:\left(\frac{n}{3}\right)\cdot12$

Then we have: .$\displaystyle \left(1\,+\,\frac{1}{\frac{n}{3}}\right)^{(\frac{n}{3})(12)}$


Let $\displaystyle z\,=\,\frac{n}{3}$. . Note that, if $\displaystyle n\to\infty$, then $\displaystyle z\to\infty$.

. . The limit becomes: . \(\displaystyle \L\lim_{z\to\infty}\left[\left(1 + \frac{1}{z}\right)^z\right]^{12}\;=\;\left[\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\right]^{12}\;=\;e^{12}\)

\(\displaystyle \L -\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\;-\)

Ha! . . . we had the same idea, galactus!
(You just dropped the "4".)