Limit of Series: x to Infinity = 0

[cscott]

With the equation $$\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}$$ can I just use the argument that 2^(x-1) will reach infinity faster than (-1)^(n+ 1) so the limit as x -> inf is 0? Because I don't see what I can do the equation to make it more "obvious".

 

[assyrian_77]

By rewriting the equation to $$-\frac{(-1)^x}{2^x}$$, it might be easier to see the solution.

 

[TD]

With the equation $$\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}$$ can I just use the argument that 2^(x-1) will reach infinity faster than (-1)^(n+ 1) so the limit as x -> inf is 0? Because I don't see what I can do the equation to make it more "obvious".

The numerator is bounded, it will always be either 1 or -1.
The denumerator, as you say, will go to infinity when x tends to infinity making the fraction tend to 0 indeed.

 

[Tide]

By the way, $$\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}$$ is not an equation! :)

 

[cscott]

By the way, $$\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}$$ is not an equation! :)

Woops!

The numerator is bounded, it will always be either 1 or -1.

Oops again

Anyway... thanks!

 

[TD]

You're welcome

 

[HallsofIvy]

And since this is a series, rather than a sequence, you might want to note that, since
$$\frac{(-1)^{x+1}}{2\cdot2^{x - 1}}= -\left(\frac{-1}{2}\right)^n$$
as you were told before, this is a geometric series.