Summation questions I got stuck on this:
(a) Find $N_1 \in \mathbb{N}_+$ such that $n \geq N_1 \implies \sum_{k=n}^\infty\frac{1}{k^2}\leq10^{-20}$
(b) Find $N_2 \in \mathbb{N}_+$ such that $n \geq N_2 \implies \sum_{k=n}^\infty\frac{1}{2^k}\leq10^{-20}$
part (b) I think I managed to do because it is a geometric series and I found that the smallest $N$ is $67$ that satisfies the condition. I can't do part (a) though, any help?
 A: It is a simple observation that:
$$ \sum_{k=n}^\infty \frac{1}{k^2} 
< \sum_{k=n}^\infty \frac{1}{k(k-1)} 
= \sum_{k=n}^\infty \left(\frac{1}{k-1} -  \frac{1}{k}\right) = \frac{1}{n-1}
$$
where the last equality follows from telescoping. So, if you want just any $N$ that will do the job, set $N = 10^{20}+1$.
In fact, the same argument used "in reverse" gives you:
$$\sum_{k=n}^\infty \frac{1}{k^2} > \sum_{k=n}^\infty \frac{1}{k(k+1)} =  \frac{1}{n},$$
so this happens to be the minimal value for $N$. Indeed, for $N' \leq N-1$ you have:
$$ \frac{1}{10^{20}} \leq \frac{1}{N'} < \sum_{k=N'}^\infty \frac{1}{k^2},$$
so any smaller value does not work.
A: The result of the first summation is PolyGamma[1,n] which is the first derivative of the Digamma function. For large values of "n", a Taylor series expansion is
PolyGamma[1,n] = 1 / n + 1 / (2 n^2) + 1 / (6 n^3) + ..
So, if we ignore the terms in n^2, n^3 (since very are very large), then, as said by other answers to this post, N = 10^20 will give you the equality. So, N = 10^20 + 1 is the answer. For such a value of N, the value of the sum is 9.9999999999999999999500*10^(-21).  
