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:

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

Is this true for non-integers?

Any relevant theorems on this?

Thanks guys

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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.

Log in to reply

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!

Log in to reply

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.

Log in to reply

Umm gp?

Log in to reply

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!!!

Log in to reply