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[SOLVED] Simplifying a summation

dwsmith

Well-known member
Feb 1, 2012
1,673
u is a complex.

Can this sum be simplified? If so, how?

$\displaystyle
\sum_{n}^{\infty}\frac{2u+5}{2}\left(-\frac{u}{2}\right)^n
$

Thanks.
 
Jan 31, 2012
54
u is a complex.

Can this sum be simplified? If so, how?

$\displaystyle
\sum_{n}^{\infty}\frac{2u+5}{2}\left(-\frac{u}{2}\right)^n
$

Thanks.


Yes.


$\sum_{n}^{\infty}\frac{2u+5}{2}(-\frac{u}{2})^n=\frac{2u+5}{2}\sum_{n}^{\infty}(-1)^n(\frac{u}{2})^n$

Can you proceed? (with $|u|<2$)
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
Yes.


$\sum_{n}^{\infty}\frac{2u+5}{2}(-\frac{u}{2})^n=\frac{2u+5}{2}\sum_{n}^{\infty}(-1)^n(\frac{u}{2})^n$

Can you proceed? (with $|u|<2$)
Proceed? I am not trying to solve just simplify it. I thought it could have been expressed a lot differently since I could have done that.

Thanks.
 
Jan 31, 2012
54
Proceed? I am not trying to solve just simplify it. I thought it could have been expressed a lot differently since I could have done that.

Thanks.


$\sum_{n}^{\infty}\frac{2u+5}{2}(-\frac{u}{2})^n=\frac{2u+5}{2}\sum_{n}^{\infty}(-1)^n(\frac{u}{2})^n=\frac{2u+5}{2}\frac{2}{u+2}=$ $\frac{2u+5}{u+2} $

(sum of geometric series)
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
$\sum_{n}^{\infty}\frac{2u+5}{2}(-\frac{u}{2})^n=\frac{2u+5}{2}\sum_{n}^{\infty}(-1)^n(\frac{u}{2})^n=\frac{2u+5}{2}\frac{2}{u+2}=\frac{2u+5}{u+2} $

(sum of geometric series)
I am not needing to solve these just simplify in the summation notation.

Can this one be simplified further?

$\sum_{n = 0}^{\infty}\left[(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)\right]$
 
Jan 31, 2012
54
I am not needing to solve these just simplify in the summation notation.

Can this one be simplified further?

$\sum_{n = 0}^{\infty}\left[(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)\right]$

$\sum_{n = 0}^{\infty}(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)=\sum_{n = 0}^{\infty}(-1)^n z^n(1-\frac{n+2}{2^n})=$

$=\sum_{n = 0}^{\infty}(-1)^n z^n-\sum_{n = 0}^{\infty}(-1)^n\frac{n+2}{2^n} z^n=$

$=\sum_{n = 0}^{\infty}(-1)^n z^n-\sum_{n = 0}^{\infty}(-1)^n n(\frac{n}{2})^n z^n-2 \sum_{n = 0}^{\infty}(-1)^n(\frac{z}{2})^n$

And each one of the sums can be evaluated.
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
$\sum_{n = 0}^{\infty}(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)=\sum_{n = 0}^{\infty}(-1)^n z^n(1-\frac{n+2}{2^n})=$

$=\sum_{n = 0}^{\infty}(-1)^n z^n-\sum_{n = 0}^{\infty}(-1)^n\frac{n+2}{2^n} z^n=$

$=\sum_{n = 0}^{\infty}(-1)^n z^n-\sum_{n = 0}^{\infty}(-1)^n n(\frac{n}{2})^n z^n-2 \sum_{n = 0}^{\infty}(-1)^n(\frac{z}{2})^n$

And each one of the sums can be evaluated.


$\sum_{n = 0}^{\infty}(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)=\sum_{n = 0}^{\infty}(-1)^n z^n(1-\frac{n+2}{2^n})$
How did you get this $\frac{n+2}{2^n}$? I can't get that.
 
Jan 31, 2012
54
$\sum_{n = 0}^{\infty}(-1)^n z^n \left(1 - \frac{1}{2^{n + 1}} - n\frac{1}{2^n}\right)=\sum_{n = 0}^{\infty}(-1)^n z^n(1-\frac{n+2}{2^n})$
How did you get this $\frac{n+2}{2^n}$? I can't get that.
Sorry! :)

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

soroban

Well-known member
Feb 2, 2012
409
Hello, dwsmith!

No one is trying "solve" anything . . . They are all trying to simplify.
Stop complaining!


$\displaystyle \text{Simplify: }\:\sum_{n=0}^{\infty}\frac{2u+5}{2}\left(-\frac{u}{2}\right)^n
$

Did you even read what A.S.Z. wrote?

He said: .$\displaystyle \sum^{\infty}_{n=0}\frac{2u+5}{2}\left(-\frac{u}{2}\right)^n \;=\;(2u+5)\underbrace{\sum^{\infty}_{n=0}\left(-\frac{u}{2}\right)^n}_{\text{geometric series}} $

We have a geometric series with: first term $a = 1$ and common ratio $r = \text{-}\frac{u}{2} $

If $ u < 2$, the series converges to a finite limit.

Can you find it?
Can you finish the problem?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hello, dwsmith!

No one is trying "solve" anything . . . They are all trying to simplify.
Stop complaining!



Did you even read what A.S.Z. wrote?

You shouldn't jump to conclusions though. You don't know what my intentions are. What is useful to my focus is only the summation not the fact it is geometric.

And CB to thank someone for something that is wrong in the first place is trite because none of you knew what I was doing with the summation. I find both of these actions rude.
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
You shouldn't jump to conclusions though. You don't know what my intentions are. What is useful to my focus is only the summation not the fact it is geometric.

And CB to thank someone for something that is wrong in the first place is trite because none of you knew what I was doing with the summation. I find both of these actions rude.
You'll have to forgive us, our psychic powers have been a little off since MHF died. We'll have to work on our mind-reading in future.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You'll have to forgive us, our psychic powers have been a little off since MHF died. We'll have to work on our mind-reading in future.
It isn't about being psychic it is about not assuming. Never assume you understand what someone else is doing because you may be wrong. Heuristics are great but not the end all be all.
 
Jan 31, 2012
54
It isn't about being psychic it is about not assuming.

But we are on math forum, board, I beg your pardon, we must assume,-that is our nature!
 

dwsmith

Well-known member
Feb 1, 2012
1,673
But we are on math forum, board, I beg your pardon, we must assume,-that is our nature!
When people are being jerks about their assumptions and making snide comments, that isn't right especially if the people making them are "distinguished" members.

I use distinguished here loosely because the acts of someone of that nature should be to the contrary.

For example, your thanking someone for an asinine remark only perpetuates the actions. Because to even speak about working on psychic powers is an indirect insult to my comment.