# Thread: [SOLVED] Finding the limit of a sequence.

1. ## [SOLVED] Finding the limit of a sequence.

I need to find the limit of {(1 + $\frac{1}{2n})^n$}, and justify my answer.

I know that $\stackrel{\lim}{n\rightarrow\infty}$ (1 + $\frac{1}{n})^n$ = e.

Can I use $\frac{n}{2}$ for my 'n,' so I get:

$\stackrel{\lim}{n\rightarrow\infty}$ (1 + $\frac{1}{2(\frac{n}{2})})^\frac{n}{2}$

= $\stackrel{\lim}{n\rightarrow\infty}$ (1 + $\frac{1}{n})^\frac{n}{2}$

= $\stackrel{\lim}{n\rightarrow\infty}$ $[(1 + \frac{1}{n})^n]^\frac{1}{2}$

= $\stackrel{\lim}{n\rightarrow\infty}$ $e^\frac{1}{2}$ = $\sqrt{e}$

Or is $\frac{n}{2}$ not a legitimate choice since I am dealing with naturals?

ilikedmath

2. That is the correct answer.
You might have seen this $\left( {1 + \frac{1}
{{2n}}} \right)^n = \left( {1 + \frac{{1/2}}
{n}} \right)^n \to e^{\frac{1}
{2}}$

3. Originally Posted by Plato
You might have seen this $\left( {1 + \frac{1}
{{2n}}} \right)^n = \left( {1 + \frac{{1/2}}
{n}} \right)^n \to e^{\frac{1}
{2}}$
Thanks, Plato! You're such a lifesaver!
It seems we're getting some spam posts on some threads here. I've noticed they're all 'newbies' too with a very few number of posts. How can we delete those spam posts?

4. Originally Posted by ilikedmath

It seems we're getting some spam posts on some threads here. I've noticed they're all 'newbies' too with a very few number of posts. How can we delete those spam posts?
Don't worry, Moderators will take care of that.

Meanwhile, if you see more spamming posts, report them.

5. Originally Posted by Krizalid
Don't worry, Moderators will take care of that.

Meanwhile, if you see more spamming posts, report them.
All right, thanks. I just found the report button. I'll report those immediately!

6. This one,