Finding a number $M$ such that $S_N<10^{-20}$ for all $N>M$

Question

Given that

$$u_n=\frac{1}{n^2-n+1} - \frac{1}{n^2+n+1}$$

$$S_n=\sum_{n=N+1}^{2N} u_n$$

Find a number $M$ such that $S_N<10^{-20}$ for all $N>M$

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I did:

$$S_N = \sum_{n=N+1}^{2N} u_n = \sum_{n=0}^{2N} u_n - \sum_{n=0}^{N} u_n$$

Using method of difference and by further simplification I get :

$$S_N= 1 - \frac{1}{(2N)^2+2N+1} - 1 + \frac{1}{N^2+N+1}$$

$$S_N= \frac{1}{N^2+N+1} - \frac{1}{4N^2+2N+1}$$

Then:

$$S_N < \frac{1}{10^{20}}$$

I did an assumption that:

$$\frac{1}{10^{20}} \approx 0$$

Because without the assumption I end up with : $4N^4+7N^3+7N^2+3N+1 > 10^{20} (3N^2+N)$ which is like way too hard

I think it's pretty good approximation for such a question

Therefore :

$$\frac{1}{N^2+N+1} < \frac{1}{4N^2+2N+1}$$

$$4N^2+2N+1 < N^2+ N+1$$

$$3N^2+N<0$$

$$\frac{-1}{3} < N < 0$$

So :

$$M=\frac{-1}{3}$$

$$\frac{3N^2+N}{(N^2+N+1)(4N^2+2N+1)} < \frac{1}{10^{20}}.$$

This is wrong. The correct answer is very large, it's:

$$M=10^{10}.$$

I have no idea how this number came, please help.

Answer 1

$M=8.7\cdot10^9$ will do. To see this, note that $4N^2+2N+1\lt4(N^2+N+1)$ hence $$S_N= \frac{1}{N^2+N+1} - \frac{1}{4N^2+2N+1}\lt\frac34\frac{1}{N^2+N+1},$$ and $S_N\lt10^{-20}$ for every $N$ such that $$N^2+N+1\gt\frac34\cdot10^{20},$$ and it suffices to ask that $$N\gt\sqrt{\frac34}\cdot10^{10}.$$ Of course, the cruder bound $10^{10}$ works, since $$S_N\lt\frac{1}{N^2+N+1}\lt\frac{1}{N^2},$$ hence $N\geqslant10^{10}$ implies $S_N\lt10^{-20}$.

Answer 2

It's much easier to see that for large $N$, the $N^2$ term dominates, so that we solve:

$$\frac{1}{N^2}-\frac{1}{4N^2}=\epsilon$$ Or: $$N=\frac{\sqrt{3}}{2\sqrt{\epsilon}}=8.66025 \times 10^9$$