Question

help!

Answer 1

what's the limit at \[+\infty\] of the function f defined by :
f(x) = \[\sum_{0}^{\infty}1\div(2^{n}+x)\]
And give an equivalent of f at \[+\infty\]

Answer 2

I found the limit but I'm stuck on the equivalent

Answer 3

the limit is 0

Answer 4

i can only say it's not 0.but i cant give the equivalent

Answer 5

why it's not 0.

Answer 6

i can show you my proof

Answer 7

To make sure I'm understanding your wording, you're asking for this, right?\[\Large
\lim_{x\to\infty}\sum_{n=0}^\infty\frac{1}{2^n+x}
\]

Answer 8

Yes!!!

Answer 9

It definitely converges, and it does appear to converge to zero, but might as well post your proof also

Answer 10

WRT the equivalence though, I am not sure.

Answer 11

Excuse me ,I ate my breakfast.

Answer 12

for my proof /
I saw that as they sought to limit \[+\infty\] well then we can restrict the study of \[\mathbb{R}+\], and we have for all x, y such that 0<=x <y is \[1\div(x+2^{n})>1\div(y+2^{n})\] by summing and taking the limit we found that:\[f(x)>f(y)\]. so f is decreasing and positive, therefore it has a limit L at + infinity. On the other hand we have for all x,
\[f(2x)=(1/2\times)f(x) +1/(1+2x)\]
So by passing to the limit at +infinity we have . L=1/2L so L=0
this is correct???

Answer 13

correcton to line 12