Estimating the Value of a Series — Question 10

Question 10 – Solution

Step 1: Reindex the tail.

Write $n=N+1+k$, where $k\ge0$. Then $$(N+1+k)^2=(N+1)^2+k(2N+2)+k^2\ge(N+1)^2+k(2N+3),$$ because $k^2\ge k$ for integers $k\ge0$.

Step 2: Compare term by term.

$$2^{-(N+1+k)^2}\le 2^{-(N+1)^2}\left(2^{-(2N+3)}\right)^k.$$ The comparison series has first term $2^{-(N+1)^2}$ and ratio $q=2^{-(2N+3)}$.

Step 3: Sum the geometric majorant.

$$\boxed{0<R_N\le\frac{2^{-(N+1)^2}}{1-2^{-(2N+3)}}}.$$ The bound is very close to the first omitted term because $q$ becomes extremely small.