Geometric Series Derivation for Given Identities

opticaltempest
Messages
135
Reaction score
0
I am trying to derive the geometric series for the following given
identities,

<br /> \begin{array}{l}<br /> \frac{1}{{0.99}} = 1.0101010101... \; \; \; {\rm{ (1)}} \\<br /><br /> \frac{1}{{0.98}} = 1.0204081632... \; \; \; {\rm{ (2)}} \\ <br /> \end{array}<br />

Here is my answer for (1),

<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{100}}} \right)} ^n + 1<br />

Here is my answer for (2),

<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{50}}} \right)} ^n + 1<br />

Are my answers correct? The only way I can get the correct answer is by
adding 1 onto the series. Is this the correct way represent the series?
 
Physics news on Phys.org
Yes, the answers are correct. The reason you are adding 1 onto the series is because this formula for the infinite sum of a geometric series
1/(1-r)
holds when the first term is 1. 1/.99 can be written as 1/(1-.01) so you have ratio r and first term 1, and the sequence you wrote for the first one reflects that only when you add 1 to it. You could sum from 0 to infinity and remove the 1, instead of from 1 to infinity and adding in the 1 afterwards, if you think it looks neater.
 
You can get rid of adding the ones by changing the summation to start from zero instad of one.
 
Ok, I see how changing the lower limit to 0 solves the problem of having
to add 1 to the series.

Thanks!
 
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...

Similar threads

Back
Top