Proof in Geometric series

[TheUserName]

Hello, I need to prove:
[math]\frac{Sn}{a1 * an}=\frac{1}{a1}+\frac{1}{a2}+\frac{1}{...}+\frac{1}{an}[/math]
Sn is the sum of the first n terms.
In one of my attempts I tried to take common factor 1 over a1 from the right side: [math]\frac{1}{a1}[\frac{1}{q}+\frac{1}{q^2}+\frac{1}{...}+\frac{1}{q^{n-1}}][/math]and a1 from the left side: [math]\frac{a1[1+q+q^2+...+q^{n-1}]}{a1*an}[/math]
I can see the similarity between the sides but I can't figure out how to continue.

 

[mmm4444bot]

I tried everything I could

Please show the start of one of your efforts, or explain why you're stuck. Thank you!

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[Steven G]

Just get your hands dirty by following the formula.

\(\displaystyle \dfrac{S_n}{a_1*a_n} = \dfrac{a_1 + a_1r + a_1r^2 + ... + a_1r^{n-1}}{a_1a_n}\)

Continue.


Regarding your work: In your 2nd expression you factored out \(\displaystyle \dfrac{1}{a_1}\). The first term you factored out \(\displaystyle \dfrac{1}{a_1}\) from was \(\displaystyle \dfrac{1}{a_1}\). You should have \(\displaystyle \dfrac{1}{a_1}( 1 + ....)\) but you have \(\displaystyle \dfrac{1}{a_1}(1/q + 1/q^2+...+ 1/q^{n-1}\). Since 1/q = 1, it must be that q=1. Then q to any power would be 1!


You can do the proof by induction. Have you considered that?

 

[TheUserName]

It's supposed to be (1 + 1/q + 1/q^2 + ... + 1/^n-1). I forgot to copy the 1.
And I'm not allowed to use induction.

 

[JeffM]

Start with the RHS.

[math]\sum_{j=0}^{n-1} \dfrac{1}{ar^{j}} = \left ( \sum_{j=0}^{n-1}\dfrac{1}{a} * \dfrac{1}{r^j} \right ) = \left \{ \sum_{j=0}^{n-1}\dfrac{1}{a} * \left ( \dfrac{1}{r}\right )^j \right \} =\\ \sum_{j=0}^{n-1} bq^j, \text { where } b = \dfrac{1}{a} \text { and } q = \dfrac{1}{r}.[/math]
See whether that helps.

 

[JeffM]

So it is now two days since the OP has checked in. This is a perfect example of where substitution of variable is so useful.

[math]\text {Given } k, \ n \in \mathbb Z^+,\ k \le n, \ t_k = ar^{(k - 1)},\ r \ne 1, \ r \ne 0, \ a \ne 0, \text { and } y = \sum_{j=1}^n \dfrac{1}{t_n}.[/math]
We can turn this into a geometric series with a pair of substitutions, namely

[math]\text {Let } b = \dfrac{1}{a} \text { and } q = \dfrac{1}{r}. \\ \text {Note } b \ne 0, \ q \ne 0, \text { and } q \ne 1.\\ \dfrac{1}{t_k} = \dfrac{1}{a * r^{(k-1)}} = \dfrac{1}{a} * \dfrac{1}{r^{(k-1)}} = \dfrac{1}{a} * \left ( \dfrac{1}{r} \right )^{(k-1)} = bq^{(k-1)}.\\ \therefore y = \sum_{j=1}^n \dfrac{1}{t_j} = \left ( \sum_{j=1}^n bq^{(1-j)} \right ) = \dfrac{b(1 - q^n}{1 - q}.\\ \therefore y = \dfrac{1}{a} * \dfrac{1 - \dfrac{1}{r^n}}{1 - \dfrac{1}{r}} = \dfrac{1}{a} * \dfrac{r^n - 1}{r^n} \div \dfrac{r - 1}{r} = \\ \dfrac{r}{ar^n} * \dfrac{r^n - 1}{r - 1} = \dfrac{1}{ar^{(n-1)}} * \dfrac{1- r^n}{1 - r} = \dfrac{1}{t_n} * \dfrac{a(1- r^n)}{a(1 - r)} =\\ \dfrac{1}{a * 1 * t_n} * \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1}{a * r^{(1-1)} * t_n} * \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1}{t_1t_n} * \dfrac{a(1 - r^n)}{1 - r} = \\ \dfrac{\displaystyle \sum_{j=1}^n a r^{(n-1)}}{t_1t_n} = \dfrac{\displaystyle \sum_{j=1}^nt_j}{t_1t_2}. \text { Q.E.D.} [/math]
Unexpected result.

 

[mmm4444bot]

it is now two days since the OP has checked in

Actually, you posted that 19 hours after the OP's previous post.

[imath]\;[/imath]

 

[Steven G]

Jeff has always been ahead of his time. Now we know exactly by how much.