Given that $$1+\frac{i^{(n)}}n=\frac{1+\frac{i^{(4)}}4}{1+\frac{i^{(5)}}5}$$ find $n$.
Your reasoning is correct. I offer you an alternative, one you might find helpful in problems similar to, but more complicated than, the one given.
Let $i^{(k)}$ be a nominal annual interest rate convertible $k$ times per year and $i$ an effective annual interest rate. If $i^{(k)}$ and $i$ are equivalent, then $${\left(1 + \frac{i^{(k)}}{k}\right)}^k = 1 + i.\tag{1}$$
Using equation $(1)$, replace each occurrence of a nominal annual interest rate by the equivalent effective annual interest rate, resulting in one equation in one variable.
$$
\begin{align*}
1 + \frac{i^{(n)}}{n} = \frac{1 + \frac{i^{(4)}}{4}}{1 + \frac{i^{(5)}}{5}}
& \implies {(1 + i)}^{\frac{1}{n}} = \frac{{(1 + i)}^{\frac{1}{4}}}{{(1 + i)}^{\frac{1}{5}}} \\
& \implies {(1 + i)}^{\frac{1}{n}} = {(1 + i)}^{\frac{1}{4} - \frac{1}{5}} \\
& \implies {(1 + i)}^{\frac{1}{n}} = {(1 + i)}^{\frac{1}{20}}.
\end{align*}
$$
Equating exponents, $\frac{1}{n} = \frac{1}{20} \implies n = 20$.
Eric deposits $X$ into a savings account at time 0, which pays interest at a nominal rate of $i$, compounded semiannually. Mike deposits $2X$ into a different savings account at time 0, which pays simple interest at an annual rate of $i$. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate $i$.
The value of Mike's investment at time $t$ is given by $$2 X (1 + ti),$$ where $t$ is measured in years. Therefore, the interest Mike earns in the last 6 months of the 8th year is given by $$2 X (1 + 8i) - 2 X (1 + 7.5i) = X i,$$ not $2 X (1 + \frac{1}{2} i)$, as covered by Ross Millikan in his answer.
As a final note, I would like to address the following:
I assume that the three nominal annual interest rates in this question are the same, although the question doesn't so specify, since if they're not then there's obviously no way to figure out this problem.
You may assume $i^{(n)}$, $i^{(4)}$, and $i^{(5)}$ are the "same," unless stated otherwise, because the variable is the same throughout. You should not assume $i^{(4)}$ and $j^{(5)}$ are the "same" because the variable changes. Note the interest rates $i^{(n)}$, $i^{(4)}$, and $i^{(5)}$ are the "same" in the sense each is equivalent to the same effective interest rate $i$ according to equation $(1)$ above, but they are not the "same" in the sense $i^{(n)} = i^{(4})$ or $i^{(4)} = i^{(5)}$ for example.
I assume you are studying for the Financial Mathematics exam. Although it has been a number of years since I sat for the actuarial exams, my fondness for actuarial science hasn't waned. I wish you the best of luck in your actuarial studies!
1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?
No. 7% semi-annual is 3.5% every six months. So annual rate is $1.035^2 - 1$.
2) Continuing with the above, if my annual nominal rate is 14%, is my annual effective rate also 14% if there is no compounding?
Yes it's 14%.
3) If I'm given a nominal rate of interest of 8% a year convertible semi-annually, what is the annual effective rate?
Is the answer to this: $(1 + .08/2)^2 = 1.0816$ --> so, effective annual rate is 8.16%?
Yes.
Why do actuaries use the term "convertible" instead of "compounded"?
Beats me.
Best Answer
The nominal interest rate is defined the way it is because, along with the compounding interval, it is a succinct way of describing how interest is computed.
If, for example, the nominal interest rate is $6$ percent, and it is compounded monthly, then we can simply divide the nominal rate by the number of months to obtain $0.5$ percent, and now we know that each month, the principal goes up by a factor of $1.005$.
The actual effective interest rate is about $6.1678$ percent, since $1.005^{12} \approx 1.0061678$, but it would be a rather ungainly way of expressing the same thing. What's more, it's likely to be only approximately correct, unless you want to carry this out to $28$ places or whatever.
To be sure, of course, we could have started with the effective interest rate, and then worked out what the nominal interest rate must be. But this requires us to compute a twelfth root, and people back in the day of hand calculators (and before that, hand computation) were understandably loath to do that. And just imagine what would happen if you were to go to daily compounding. (In many ways, compounding continuously is easier, though it requires taking a logarithm.) It was simply easier to deal with the nominal interest rate.
Also, from a marketing perspective, it was easier to tell people that their effective rate was higher than their nominal rate (sounds like they're getting a compounding bonus) than that the rate they actually got each month was less than the effective rate divided by $12$ (sounds like compounding costs them money).