Am I doing this series right? (arithmetic question but calc)

[Rijad Hadzic]

Homework Statement

I have series

$\sum_{n=1}^\infty (1/n)(2^n)(-1/2)^n$

Homework Equations

The Attempt at a Solution

So trying to do the solution

$(1/n)(2^n)(-\frac {1^n}{2^n})$

since 1^n is going to be one for all values of n, can I say,

$(1/n)(2^n)(-\frac {1}{2^n})$

then

$(1/n)(\frac{2^n}{-2^n})$

then that gives me
$(1/n)(-1)$

and I know 1/n is divergent, but this answer is wrong? The reason I think its wrong is because I can't consider 1^n as just one, correct? But even if I do still consider $1^n$ as $1^n$ instead of just $1$, my answer is $(1/n)(-1)(1^n)$ which is different from $(1/n)(-1^n)$

If I go

$(1/n)(\frac{2^n}{-2^n}) (1^n)$

I get

$(1/n)(-1)(1^n)$

But this is still divergent.

If I go from

$(1/n)(\frac{2^n}{2^n}) (-1^n)$

I get $(1/n)(1)(-1^n)$

which is convergent according to the alternating series test which is the answer.

I just don't understand when I have

$(1/n)(\frac{2^n}{-2^n}) (1^n)$

how I don't get the same answer, because I get

$(1/n)(-1) (1^n)$

which is different from

$(1/n)(-1^n)$

why would it matter if you take the negative sign with the denominator or the numerator?

 

[Rijad Hadzic]

Or if I'm here

$\sum_{n=1}^\infty (1/n)(2^n)(\frac {-1}{2})^n$

then go

$\sum_{n=1}^\infty (1/n)(2^n)(\frac {-1^n}{2^n})$

and multiply straight out

$\sum_{n=1}^\infty (1/n)(\frac {(2^n)(-1^n)}{2^n})$

I get $\sum_{n=1}^\infty (1/n)(-1)^n$ which is convergent.

I just don't understand why I can't go from $(1/2)^n$ to $(1^n) * (1/2^n)$

for example, if I go from

$\sum_{n=1}^\infty (1/n)(\frac {(2^n)(1^n)}{-2^n})$

to

$\sum_{n=1}^\infty (1/n)(-(1^n))$

I get a completely different answer.

One alternates from negative to positive, the other is negative all the time...
How am I suppose to know if the negative is on the bottom or top??

 

[scottdave]

Homework Statement

I have series

$\sum_{n=1}^\infty (1/n)(2^n)(-1/2)^n$

Homework Equations

The Attempt at a Solution

So trying to do the solution

$(1/n)(2^n)(-\frac {1^n}{2^n})$

since 1^n is going to be one for all values of n, can I say,

$(1/n)(2^n)(-\frac {1}{2^n})$

then

$(1/n)(\frac{2^n}{-2^n})$

then that gives me
$(1/n)(-1)$

and I know 1/n is divergent...
why would it matter if you take the negative sign with the denominator or the numerator?

Your problem comes from assuming that $\frac {2^n} {-2^n}$ will cancel out to be -1 for all values of n.
When n is even, the denominator is positive, so you have 1, and when n is odd you do get -1, so it alternates between 1 and -1.

 

[Rijad Hadzic]

Your problem comes from assuming that $\frac {2^n} {-2^n}$ will cancel out to be -1 for all values of n.
When n is even, the denominator is positive, so you have 1, and when n is odd you do get -1, so it alternates between 1 and -1.

Thanks a lot bro. That makes so much sense.

 

[scottdave]

Just take care to watch the parentheses.

 

[Mark44]

Homework Statement

I have series

$\sum_{n=1}^\infty (1/n)(2^n)(-1/2)^n$

You're making this harder than it needs to be. $2^n(-1/2)^2 = 2^n(1/2)^n(-1)^n = (-1)^n$.
I'm using the fact that $(ab)^n = a^n b^n$ to write $(-1/2)^2$ as $(-1)^n$ times $(1/2)^n$.

 

[jishnu]

Could someone please explain me what is this [itex] written almost every where in this question, because of that I am unable to understand the question

 

[Mark44]

Could someone please explain me what is this [itex] written almost every where in this question, because of that I am unable to understand the question

These are LaTeX tags. Under ordinary circumstances, you shouldn't be seeing them, as the browser uses them to display mathematics symbols.

See our LaTeX tutorial here -- https://www.physicsforums.com/help/latexhelp/

 

[jishnu]

These are LaTeX tags. Under ordinary circumstances, you shouldn't be seeing them, as the browser uses them to display mathematics symbols.

See our LaTeX tutorial here -- https://www.physicsforums.com/help/latexhelp/

Thanks for the assistance, this seems a bit complicated for a beginner like me but hoping to cop up with new language (for me) soon... :)