Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.

How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?

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- Oct 23rd 2006, 11:42 AMJimmyTLimit of e
Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.

How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it? - Oct 23rd 2006, 12:46 PMearboth
Hi,

this one looks like a trick - but it works:

$\displaystyle \left(1+\frac{2}{n}\right)^n$ = $\displaystyle \left(1+\frac{1}{\frac{n}{2}}\right)^n$ =

$\displaystyle \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2$

Now substitute n/2 = y. Then you get:

$\displaystyle \left( \left(1+\frac{1}{y}\right)^y\right)^2$

$\displaystyle \lim_{y\rightarrow \infty}{\left(1+\frac{1}{y}\right)^y}=e$

Therefore $\displaystyle \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2=e^2$

EB - Oct 23rd 2006, 12:52 PMCaptainBlack
- Oct 23rd 2006, 04:25 PMJimmyT
Along the same lines, how would you manipulate the lim (n to inf) of

(3n)^(1/(2n)) to get 1?

Jim - Oct 23rd 2006, 04:50 PMThePerfectHacker
- Oct 23rd 2006, 04:57 PMThePerfectHacker
I am not going to write infinity because all limits of sequences are taken long infinite.

You have,

$\displaystyle \lim (3n)^{\frac{1}{2n}}=\lim 3^{\frac{1}{2n}}n^{\frac{1}{2n}}=\lim \sqrt{3}^{1/n}\cdot \sqrt{n}^{1/n}$

We divide the problem into two limit problems,

1)$\displaystyle \lim \sqrt{3}^{1/n}=\sqrt{3}^0=1$ (sequence composition rule).

2)$\displaystyle 1\leq \sqrt{n}^{1/n}\leq n^{1/n}$

Squeeze theorem, the limit,

$\displaystyle \lim n^{1/n}=1$ is well known.

Thus the answer is $\displaystyle 1\cdot 1 =1$