# Thread: limit of a sequence

1. ## limit of a sequence

Hey everyone.
I wanted to get help finding the limit of the following sequence (n->inf):
lim(1/(2^n + 1) + 2/(2^n + 2) + 4/(2^n + 3) + .. + 2^(n-1)/(2^n + n))

I am having trouble because the number of operands to add depends on n.

Thanks a lot,
Michael.

2. ## Re: limit of a sequence

Is this the question you have?
$\lim_{n\rightarrow\infty} \frac{1}{2^{n+1}} + \frac{2}{2^{n+2}} + \frac{4}{2^{n+3}} + ... + \frac{2^{n-1}}{2^{n+n}}$

3. ## Re: limit of a sequence

Exactly, Thanks

4. ## Re: limit of a sequence

is the last term right, from the pattern I see it seems there's a typo for the last term?

EDIT: Nevermind!

5. ## Re: limit of a sequence

Okay so the nth term is $a_{n} = \frac{2^{n-1}}{2^{2n}}$ which simplifies to
$a_{n} = \frac{1}{2^{n-1}}$

so you have $\lim_{n\rightarrow\infty}a_{n}$

Do you think you can do it now?

6. ## Re: limit of a sequence

I just took another look at the question and it seem I made a mistake.
The +n in the denominator is not for the power, it is just +n.
So the nth term would be 2^(n-1)/((2^n) + n)