Yearly discounting to equivalent monthly discounting Say annual benefit is $120$. Taking an annual discount rate of 3.5%, the present discounted value of the benefit a year from now is $\frac{120}{1+0.035}=115.94$. Assume that benefit is accrued evenly throughout the year i.e. $10$ per month. 
How do I go about calculating a monthly discount rate such that the sum of the discounted monthly benefits is $115.94$ after a year?
EDIT:
To clarify, I would like $r$ such that $\sum_{n=1}^{12} \frac{10}{(1+r)^n} = 115.94$.
Generally, is there some way to convert from a yearly discount  rate to a monthly one? 
 A: 
I would like $r$ such that $\sum_{n=1}^{12} \frac{10}{(1+r)^n}=115.94$

This can be managed with the partial sum of the geometric series. Let $q=1+r$ and $S_n=115.94$
$\sum\limits_{n=1}^{12} \frac{10}{(1+r)^n}$
$=S_n=10\cdot \left( \ \   \quad \ \frac1{q}+ \frac1{q^2}+ \frac1{q^3}+ \ldots + \frac1{q^{11}}+ \frac1{q^{12}}\right) \qquad (1)$
$q\cdot S_n=10\cdot\left(1+\frac1{q}+ \frac1{q^2}+ \frac1{q^3}+ \ldots + \frac1{q^{11}}\right)\qquad (2)$ 
Subtracting (2) from (1).
$S_n\cdot (1-q)=10\cdot \left(\frac1{q^{12}}-1 \right)$
$S_n\cdot (q^{12}-q^{13})=10\cdot \left(1-q^{12} \right)$
$-S_n\cdot q^{13}+(S_n+10)\cdot q^{12}-10=0$
This is a polynomial with a degree of 13. Unfortunately this equation cannot be solved algebraically. Thus you have to solve the equation numerically. For this purpose you can apply the Newton-Raphson method. 
A: Well, you have 12 cash flows, each a month from the previous one. So your portfolio looks like 10 next month, and 10 the following month, etc. If you assume the same monthly discounting rate $r$, can you find the value of the portfolio $V(r)$ as a function of $r$?
HINT (Find the value for each payment and add them together)
Then, solve $V(r) = 115.94$.
