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 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 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&#039;t be seeing them, as the browser uses them to display mathematics symbols.<br /> <br /> See our LaTeX tutorial here -- <a href="https://www.physicsforums.com/help/latexhelp/" class="link link--internal">https://www.physicsforums.com/help/latexhelp/</a>

 

[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... :)