Summation of series with terms $U_n=\frac{1}{n^2-n+1} -\frac{1}{n^2+n+1}$

Given that $U_n=\dfrac{1}{n^2-n+1} -\dfrac{1}{n^2+n+1}$, find $S_N$= $\sum_{n=N+1}^{2N}U_n$ in terms of $N$. Find a number $M$ such that $S_n<10^{-20}$ for all $N>M$.

I was able to calculate the sum as $\dfrac{1}{N^2+N+1}-\dfrac{1}{4N^2+2N+1}$ using the method of differences. I am having trouble doing the second part. After thinking a while I think that the correct approach is to remove some of the terms while preserving the inequality and then solve for $N$. Two such attempts were $\dfrac{1}{N^2+N+1}-\dfrac{1}{N^2}$ and $\dfrac{1}{N^2+N+1}-\dfrac{1}{N}$ but I don't know how to proceed. Is this the correct way? Any help, including hints, would be appreciated

You're overthinking it. Just use some crude bounds: \begin{align*} S_N=\frac{1}{N^2+N+1}-\frac{1}{4N^2+2N+1}&< \frac{1}{N^2+N+1}\\ &< \frac{1}{N^2}. \end{align*} Therefore $M=10^{10}$ works, by which we mean that for any $N>10^{10}$, $S_N<10^{-20}$: $$S_N<\frac{1}{N^2}<\frac{1}{10^{20}}$$

• Does this mean that M can take on different values? Like $10^{20}$ from $\frac{1}{n}$? – user140161 Dec 31 '14 at 20:22
• $M$ is just the constant number $10^{10}$. Then $S_N<10^{-20}$ whenever $N>10^{10}$. Does that make sense? – pre-kidney Dec 31 '14 at 20:30
• Yes but what I'm asking is, can't $M$ be $10^{20}$ too? Because in that case too the condition for all $N>M$, $S_N<10^{-20}$ is satisfied. – user140161 Dec 31 '14 at 22:38
• Yes, then $S_N<10^{-40}$. In general, any $M$ larger than $10^{10}$ will work (but if you go smaller, you will run into issues). – pre-kidney Jan 3 '15 at 1:42

$$\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}=\\\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}=\\f(n)-f(n+1)$$ $$\sum_{n=N+1}^{2N}(f(n)-f(n+1))=f(N+1)-f(2N)$$

• Did you read the question? Computing the value of $S_N$ is not what is asked. – Did Jan 11 '15 at 13:58
• To echo Did's comment, the asker indicated in the question that he had already computed the closed form of $S_N$, correctly. The actual question here is how to find $M$. It's sad how many answerers and upvoters failed to carefully read the question... – epimorphic Jan 13 '15 at 20:37