1. ## Convergent Series

First time poster, forgive me in advance.

I am trying to get a function out of the following series that goes on for infinity. I'm pretty sure this is possible as I was able to solve the problem using Excel but I'm stumped when trying to assign a function to it that can be done by hand.

The series is:

1.1/1.12 + 1.2/1.12^2 + 1.3/1.12^3 + 1.4/1.12^4 + ... to infinity

When I solve this in Excel I can get a value of 16.11 however I can't seem to get this number by hand. Is it possible to do by hand, and if so, how?

Please let me know if I need to clarify anything.

2. Originally Posted by ShoelessJoe
The series is:

1.1/1.12 + 1.2/1.12^2 + 1.3/1.12^3 + 1.4/1.12^4 + ... to infinity

When I solve this in Excel I can get a value of 16.11 however I can't seem to get this number by hand. Is it possible to do by hand, and if so, how?
$\displaystyle \sum_{n=1}^{\infty} \frac{1+\frac{n}{10}}{ (1.12)^n} = \sum_{n=1}^{\infty} \frac{1}{(1.12)^n} + \frac{1}{10} \sum_{n=1}^{\infty} n\left( \frac{1}{1.12} \right)^n$

The first sum is geometric. The second sum can be obtained from the geometric. Note, $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \mbox{ for }x\in (-1,1)$. Differenciate to get, $\displaystyle \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \mbox{ for }x\in (-1,1)$. Thus, $\displaystyle \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \mbox{ for }x\in (-1,1)$.

3. Hello, ShoelessJoe!

Welcome aboard!

Evaluate: .$\displaystyle \frac{1.1}{1.12} + \frac{1.2}{1.12^2} + \frac{1.3}{1.12^3} + \frac{1.4}{1.12^4} + \frac{1.5}{1.12^5} + \hdots$

We are given the series: .$\displaystyle S \;=\;\frac{1.1}{1.12} + \frac{1.2}{1.12^2} + \frac{1.3}{1.12^3} + \frac{1.4}{1.12^4} + \frac{1.5}{1.12^5} + \hdots$

Multiply by $\displaystyle \frac{1}{1.12}\!:\qquad\quad\frac{1}{1.12}S \;=\qquad\quad\;\frac{1.1}{1.12^2} + \frac{1.2}{1.12^3} + \frac{1.3}{1.12^4} + \frac{1.4}{1.12^5} + \hdots$

Subtract: . . . . $\displaystyle \left(1 - \frac{1}{1.12}\right)S \;=\;\frac{1.1}{1.12} + \frac{0.1}{1.12^2} + \frac{0.1}{1.12^3} + \frac{0.1}{1.12^4} + \frac{0.1}{1.12^5} + \hdots$

And we have: . . . . . $\displaystyle \frac{0.12}{1.12}S \;=\;\frac{1.1}{1.12} + \frac{0.1}{1.12^2}\underbrace{\left(1 + \frac{1}{1.12} + \frac{1}{1.12^2} + \frac{1}{1.12^3} + \hdots\right)}_{\text{geometric series}}$

The geometric series has first term $\displaystyle a = 1$ and common ratio $\displaystyle r = \frac{1}{1.12}$
. . Its sum is: .$\displaystyle \frac{1}{1 - \frac{1}{1.12}} \:=\:\frac{1.12}{0.12}$

Then we have: .$\displaystyle \frac{0.12}{1.12}S \;=\;\frac{1.1}{1.12} + \frac{0.1}{1.12^2}\left(\frac{1.12}{0.12}\right) \;=\;\frac{0.232}{(0.12)(1.12)}$

Therefore: .$\displaystyle S \;=\;\frac{1.12}{0.12}\cdot\frac{0.232}{(0.12)(1.1 2)} \;=\;\frac{0.232}{0.12^2} \;=\;16.11111\hdots\;=\;\frac{145}{9}$

Please check for egregious blunders . . .
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