
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^(n1)/(2^n + n))
I am having trouble because the number of operands to add depends on n.
Thanks a lot,
Michael.

Re: limit of a sequence
Is this the question you have?
$\displaystyle \lim_{n\rightarrow\infty} \frac{1}{2^{n+1}} + \frac{2}{2^{n+2}} + \frac{4}{2^{n+3}} + ... + \frac{2^{n1}}{2^{n+n}}$

Re: limit of a sequence

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!

Re: limit of a sequence
Okay so the nth term is $\displaystyle a_{n} = \frac{2^{n1}}{2^{2n}}$ which simplifies to
$\displaystyle a_{n} = \frac{1}{2^{n1}}$
so you have $\displaystyle \lim_{n\rightarrow\infty}a_{n}$
Do you think you can do it now?

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^(n1)/((2^n) + n)