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- Thread starter dwsmith
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- Jan 31, 2012

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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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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.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 $|u|<2$)

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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I am not needing to solve these just simplify in the summation notation.$\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)

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

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

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

And each one of the sums can be evaluated.

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

- Jan 31, 2012

- 54

Sorry!How did you get this $\frac{n+2}{2^n}$? I can't get that.

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

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

No one is trying "solve" anything . . . They are all trying to

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?

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- #10

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.Hello, dwsmith!

No one is trying "solve" anything . . . They are all trying tosimplify.

Stop complaining!

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

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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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.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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- #12

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.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.

- Jan 31, 2012

- 54

It isn't about being psychic it is aboutnot assuming.

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

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- #14

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

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.