I am having trouble deciding which of the expression is larger. The following is the original problem and I may not have the expression entirely correct, but I am pretty confident.
A loan of $L$ is being repaid by $n$ payments which starts one period after the loan was made. The effective interest is $i$ per period. There are two schemes:
Scheme A), Each payments are level.
Scheme B), Each payments equal $\frac{L}{n}$ plus the interest due from the previous outstanding balance.
Show that the total amount of interest paid in scheme A is greater than that of B.
Here is my claim.
Under scheme A
$$L = K(v+v^2+ \cdots + v^n) = Ka_{\overline{n}\rceil i}$$
and the sum of the interest can be calculated as
$$I_{TA}=K((1-v)+(1-v^2)+ \cdots +(1-v^n))=K(n-a_{\overline{n}\rceil i})$$
therefore,
$$I_{TA}=\frac{L}{a_{\overline{n}\rceil i}}(n-a_{\overline{n}\rceil i})$$
Now for scheme B, the principal paid is always $\frac{L}{n}$ which makes the outstanding balance
$$OB_t=L(1-\frac{t}{n})$$
The interest of these values add up to
$$\begin{align} I_{TB} &= i(OB_0+OB_1+ \cdots + OB_{n-1})\\ &=Li(1+(1+\frac{1}{n})+ \cdots + (1-\frac{n-1}{n})+(1-\frac{n}{n}))\\ &=Li(n-\frac{1}{n}(\frac{n(n+1)}{2}))\\ &=Li\frac{n-1}{2}\\ \end{align}$$
So my goal is to prove that $$I_{TA} \ge I_{TB}$$ which is intuitively clear, but I am having trouble algebraically showing it. Also, I would assume that this problem assumes a natural situation where $n \in \mathbb N$, $i > 0$, etc.
It isn't pretty, but would you agree for $i>0 , n \in \mathbb{N}$ that $\underbrace{\tfrac{i}{2}\left(n+1\right)}_{>0} \geq \underbrace{1-n}_{\leq 0}$?
If so you can use (lots of) algebra to achieve $I_{TA} \geq I_{TB}$ the desired result. It took me about 10 steps with doing some one or two manipulations per step. Assuming I did not make any errors and typed this correctly, some key steps include:
$$\dfrac{i}{2}\left(n+1\right)\geq 1-n$$ $$\dfrac{ni}{d}- \dfrac{ni}{2} + \dfrac{i}{2}\geq 1$$ $$\dfrac{ni}{1-v^n} \geq 1 - \dfrac{i}{2}(n-1)$$ $$n \geq a_{\overline{n}|} \left( 1 + \dfrac{i}{2}(n-1)\right)$$ $$\dfrac{n - a_{\overline{n}|}}{a_{\overline{n}|}} \geq \dfrac{i}{2}(n-1)$$ $$I_{TA} \geq I_{TB}$$