What is the Convergent Series Sum Formula?

jediwhelan
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Homework Statement



Dear All,

I have a series that I know to converge but for which I can't work out the infinite sum. It should be something simple.

<br /> S_n = \sum_{j=1}^\infty \beta^j j<br />

Can somebody help me with this?

I think the solution is:

<br /> \frac{\beta}{(1-\beta)^2}<br />
 
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It looks right:

S_n = \beta + 2\beta^2 + 3\beta^3 + \cdots

\frac{S_n}{\beta} = 1 + 2\beta + 3\beta^2 + \cdots

(\frac{1}{\beta}-1) S_n = 1 + \beta + \beta^2 + \cdots

(\frac{1}{\beta}-1) S_n = \frac{1}{1-\beta}

(\frac{1-\beta}{\beta}) S_n = \frac{1}{1-\beta}

S_n = \frac{\beta}{(1-\beta)^2}

It's called an arithmetico-geometric series I think,

\displaystyle\sum_{n=0}^{\infty}(a+nd)r^n = \frac{a}{1-r} + \frac{rd}{(1-r)^2}
 
brilliant. I was trying something like that but couldn't get it.

Thanks for the quick reply.

Paul
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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