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Something I realized about powers:

I was walking around (currently on vacation) and I realized something which I think is true (and I'm sure have been thought of before):

For all positive integer \(b > 1\)...

\( \sum_{n = 0}^k b^n = \frac {b^{n+1} - 1}{b - 1}\)

So I ask a few questions:

  1. Can somebody prove this? I'm only a high school student so I'm not totally sure how to do this.

  2. Is this true for non-integers?

  3. Any relevant theorems on this?

Thanks guys

Note by Michael Tong
4 years, 3 months ago

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2 votes

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Is not this, the sum of a geometric progression? I am sure you already know how to get this formula.

Aditya Parson - 4 years, 3 months ago

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The series is nothing but a G.P. Take the sum as S and multiply S by b, now subtract S from bS and you will get the result. Simple!!!

Siddharth Kumar - 4 years, 3 months ago

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Umm gp?

Soham Chanda - 4 years, 3 months ago

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S = 1 + b + b^2 + b^3 +....+ b^n

Multiply b both sides. We have two equations -

S = 1 + b + b^2 + b^3 +....+ b^n

b*S = b + b^2 + b^3 + .....+ b^n+1

Subtract the two equations. You will get -

S(b - 1) = b^n+1 - 1

Multiply (b - 1) both sides and there you go!

Lokesh Sharma - 4 years, 3 months ago

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One more, since this proof is valid for every real number except 1 I suppose, this theorem works for all real numbers except 1. I am not 100% sure though. Please tell me if there are some other numbers other than 1.

Lokesh Sharma - 4 years, 3 months ago

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Theorem: The series \(\sum a_n\) converges if \[\lim_{n\to \infty} \sup \sqrt[n]{}{\left |a_n \right |<1.}\]

Definition:

http://en.wikipedia.org/wiki/Supremum

Let \({s_n}\) be a sequence of real numbers. Let \(E\) be the set of numbers x such that \(s_{n_k} \to x\) for some subsequence \({s_{n_k}}\). Put \[\lim \sup x_n= \sup E.\]

Proof of the Theorem:

If \[\alpha=\lim_{n\to \infty} \sup \sqrt[n]{\left |a_n \right |}\] then there exists \(\beta >0\) such that \(\alpha< \beta<1\). Then there there exists \(N \geq 0\) such that \(N \leq n\) implies \[\left |a_n \right |\leq \beta ^n\]. Then the convergence of geometric series implies the convergence of this series here.

From this theorem, you can evaluate 'radius of convergence' for a complex power series.

Okay Nho - 4 years, 3 months ago

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