Savings question: How long will savings last if I withdraw a certain amount every year? Let's say that for $30$ years I insert $.20$ dollars every year into a bank account for with interest rate $5\%$. After 30 years, I stop inserting money, and start withdrawing 0.70 dollars every year. Question: For how many years will I be able to withdraw 0.7 dollars before the account is empty?
My approach is by calculating the amount saved after $30$ years: I get $0.2 \times (\sum_{i=1}^{30} (1.05)^i)$. Then, we start withdrawing 0.7 dollars from here on out for, say, $s$ years, and by looking at the calculations for each year, I generalize to the following formula for the bank account after $30 + s$ years: $0.2 \times (\sum_{i=1}^{30+s} 1.05^i) - 0.7 \times (\sum_{i=1}^{s} 1.05^i) $. 
Is this correct so far? And what next? I want the $s$ which makes the bank account zero, but solving that equation in Maple gives me complex numbers. What's gone wrong?
 A: Let be $P=0.2$, $Q=0.7$, $n=30$, $i=5\%$. Let suppose that deposits and withdrawals are made at the beginning of each year.
After $n$ year we will have the value $F$
$$
F=P\sum_{k=1}^n(1+i)^k=P\,\ddot s_{\overline{n}|i}
$$
From $n+1$ to $n+x$ we will withdraw $Q$ and then the final value of the withdrawals after $x$ years will be
$$
E=Q\sum_{k=1}^{x}(1+i)^k=Q\,\ddot s_{\overline{x}|i}
$$
We have to find $x$ such that
$$
F(1+i)^x-E=0
$$
that is
$$
P\,\ddot s_{\overline{n}|i}(1+i)^x=Q\,\ddot s_{\overline{x}|i}
$$
and observing that $\ddot s_{\overline{x}|i}=(1+i)^x\,\ddot a_{\overline{x}|i}$
$$
P\,\ddot s_{\overline{n}|i}=Q\,\ddot a_{\overline{x}|i}
$$
and that means that the future value of the deposits must be equal to the present value of the withdrawals. Obseving that $\ddot s_{\overline{n}|i}=(1+i)\,s_{\overline{n}|i}$ and $\ddot a_{\overline{x}|i}=(1+i)\,a_{\overline{x}|i}$ and solving for $x$
$$
\tfrac{P}{Q}\,i\,s_{\overline{n}|i}={1-v^x}\qquad\Longrightarrow\quad x=\frac{\log\left(1-\frac{P}{Q}\,i\,s_{\overline{n}|i}\right)}{\log v}= 61.04532 \approx 61
$$
A: I think $0.2 \times (\sum_{i=1}^{30+s} 1.05^i)$ would only be correct if you continued to deposit 0.2 \$ for $s$ years after the 30 years. If you stop, then it should be $0.2 \times (\sum_{i=1}^{30} 1.05^i) \times 1.05^s$. Let $a$ be the amount of money after 30 years, i.e. $a = 0.2 \times (\sum_{i=1}^{30}1.05^i)$. Then you withdraw 0.7 \$, so after 31 years you'd have $(a-0.7)\cdot 1.05 = a\cdot 1.05 - 0.7\cdot 1.05$. By the same procedure, after 32 years you'd have $((a\cdot 1.05 - 0.7\cdot 1.05) - 0.7)\cdot 1.05 = a\cdot 1.05^2 - 0.7\cdot 1.05^2 - 0.7\cdot 1.05$. So if my reasoning is correct, after $s$ years you'd have $a\cdot 1.05^s - 0.7\cdot(\sum_{i=1}^{s} 1.05^i)$.
A: WLOG, fast forward to $30$ years later. You have been depositing $0.2$ dollars per year for the last $30$ years. today you start the first of $s$ yearly withdrawals of $0.70$ dollars until the balance drops to zero in the $(s-1)$-th year.
Equating the cumulative compounted value today of past deposits to the present value of $s$ withdrawals of $0.70$:
$$\begin{align}
0.2\sum_{n=1}^{30} 1.05^n&=0.7\sum_{n=0}^{s-1}\frac 1{1.05^n}\\\\
0.2\left[\frac{1.05}{0.05} \bigg(1.05^{30}-1\bigg)\right]&=\frac {0.7}{0.05}\left(1.05-\frac 1{1.05^{s-1}}\right)\\\\
\underbrace{\frac{0.2}{0.7}\bigg(1.05^{30}-1\bigg)}_{0.949126}&=1-\frac 1{1.05^s}\\
1.05^s&=\frac1{1-0.949126}=19.656\\\\
s&=\frac{\log 19.656}{\log 1.05}\approx 61\quad\blacksquare
\end{align}$$
