# Limit of Series

1. Oct 23, 2013

### elementbrdr

Hi,

I don't know how to analyze the following, but I am wondering whether there is a way to determine whether a series of the following form is convergent: V$_{n}$=(V$_{n-1}$+a)/b. Thank you.

2. Oct 23, 2013

### jbriggs444

A series can only be convergent (in the usual sense) if the limit of its terms is zero. If a is non-zero, what effect does that have on the limit of the terms in the series?

3. Oct 23, 2013

### pasmith

This is a linear recurrence relation:

$$V_n - \frac{V_{n-1}}{b} = \frac{a}{b}$$

For $b \neq 1$ the solution is $$V_n = \frac{A}{b^n} + \frac{a}{b-1}$$ for an arbitrary constant $A$. Thus it will converge if and only if $|b| > 1$ or $A = 0$ and its limit will be $\frac{a}{b-1}$.

If $b = 1$ then the solution is $$V_n = A + na$$ for an arbitrary constant $A$, and it does not converge unless $a = 0$.

4. Oct 23, 2013

### elementbrdr

Thank you. I see your points. Maybe, in framing this as a limit of a series, I am thinking about the underlying problem incorrectly. So I will expand on the problem I am trying to solve. In a financial context, I am trying to calculate a return that is inclusive of a return-based payment. More specifically, I am trying to calculate a payment (V), which payment is equal to a constant times return, i.e., b*(R/C -1), where R equals final value and C equals initial value. However, final value (R) is defined to include V, such that R= V+a. So the definition is circular. Intuitively, I thought of this as a limit of a series, building inward, where in the first step, the expression would be V = b*(R/C-1) = b*((V+a)/C-1); and in the second step, the expression would be V = b*(((b*(R/C-1))+a)/C-1) = b*(((b*((V+a)/C-1))+a)/C-1). So on and so forth. I would appreciate any guidance as to how to approach this problem. Thanks in advance.

5. Oct 24, 2013

### HallsofIvy

6. Oct 24, 2013

### elementbrdr

I think that what I am trying to do falls within the category of a sequence, rather than a series. I am interested in what the value of the expression would be after an infinite number of iterations of the "steps" illustrated above (and not in the sum of the expressions produced by each step). Sorry to have started off the thread with a misconception.

7. Oct 25, 2013

### elementbrdr

Viewing my question in terms of a sequence rather than a series, is there a way to state the calculation I described as a simple expression? I would be happy to try to clarify further if my question is still muddled. Thank you.