April 16th
Today I learned about Kummer series acceleration, from Keith Conrad of course. This is the idea.
Definition. Given $\{a_k\}_{k=1}^\infty$ whose sum absolutely converges, if we are given a sequence $\{b_k\}_{k=1}^\infty$ such that $\frac{a_k}{b_k}\to\lambda$ and we know $B=\sum b_k,$ then we note \[\sum_{k=1}^\infty a_k=\lambda B+\sum_{k=1}^\infty(a_k-\lambda b_k)=\lambda B+\sum_{k=1}^\infty\left(1-\frac{b_k}{a_k}\right)a_k,\] which has good chances of converging faster.
We note that the absolutely convergent condition on $\{a_k\}_{k=1}^\infty$ is really there to let us extract out $\lambda B$ in the rudest way possible. In practice, I think this method of series acceleration is done when the terms are positive mostly.
It may seem like a significant roadblock to find the sequence $\{b_k\}_{k=1}^\infty,$ but we can actually cover all polynomial growth pretty well.
Lemma. Given some positive integer $d,$ we have that \[\sum_{k=1}^\infty\frac1{k(k+1)\cdots(k+d)}=\frac1{d\cdot d!}.\]
We observe that\[\frac{1/d}{k(k+1)\cdots(k+d)}-\frac{1/d}{(k+1)(k+2)\cdots(k+d+1)}=\frac{(k+d)/d-k/d}{k(k+1)\cdots(k+d+1)},\]and the right-hand side is the desired summand $\frac1{k(k+1)\cdots(k+d)}.$ It follows that\[S_d:=\sum_{k=1}^\infty\frac1{k(k+1)\cdots(k+d+1)}=\frac1d\sum_{k=1}^\infty\left(\frac1{k(k+1)\cdots(k+d)}-\frac1{(k+1)\cdots(k+d+1)}\right).\]Shifting (not rearranging) the sum, this gives\[S_d=\frac1d\cdot\frac1{1\cdot2\cdots d}+\sum_{k=2}^\infty\left(-\frac1{(k+1)\cdots(k+d+1)}+\frac1{(k+1)\cdots(k+d)}\right).\]As promised, this does evaluate to $\frac1{d\cdot d!},$ so we are done. $\blacksquare$
As an example, we accelerate $\zeta(3)\approx1.20205690315.$ The standard sum is\[\zeta(3)=\sum_{k=1}^\infty\frac1{k^3},\]but we can accelerate the convergence of this. After computing $N$ terms, we can bound the error as\[\zeta(3)-\sum_{k=1}^N=\sum_{k=N+1}^\infty\frac1{k^3} \lt \int_N^\infty\frac1{x^3}\,dx=\frac1{2N^2}\]by viewing the sum as a Riemann sum. So, for example, $10$ terms gives $\zeta(3)\approx1.19753198567.$
Now, from the above lemma, we use $\frac1{k(k+1)(k+2)}$ with $\frac{k(k+1)(k+2)}{k^3}\to1.$ The lemma tells us that the auxiliary sum is $\frac14,$ and the leftover term is\[\frac1{k^3}-\frac1{k(k+1)(k+2)}=\frac{3k+2}{k^3(k+1)(k+2)},\]which we can see decays about a degree faster. Anyways, our new sum is\[\zeta(3)=\frac14+\sum_{k=1}^\infty\frac{3k+2}{k^3(k+1)(k+2)}.\]Again, after computing $N$ terms, we can bound the error as\[\sum_{k=N+1}^\infty\frac{3k+2}{k^3(k+1)(k+2)} \lt \sum_{k=N+1}^\infty\frac{3(k+1)}{k^3(k+1)k} \lt \int_N^\infty\frac3{k^4}\,dx=\frac1{N^3}.\]Explicitly, we can see this with $\zeta(3)\approx1.20131986446$ after merely $10$ terms, which is much better.
The real point of series acceleration is that we can do this multiple times. Our current sequence grows $\Theta\left(3/k^4\right),$ so we can choose $\frac3{k(k+1)(k+2)(k+3)}$ as our auxiliary sequence, which has sum $\frac16.$ The term now looks like\[\frac{3k+2}{k^3(k+1)(k+2)}-\frac3{k(k+1)(k+2)(k+3)}=\frac{(3k+2)(k+3)-3k^2}{k^3(k+1)(k+2)(k+3)}.\]So our sum is now\[\zeta(3)=\frac14+\frac16+\sum_{k=1}^\infty\frac{11k+6}{k^3(k+1)(k+2)(k+3)}.\]And after computing $N$ terms, the error is bounded by\[\sum_{k=N+1}^\infty\frac{11k+6}{k^3(k+1)(k+2)(k+3)} \lt \sum_{k=N+1}^\infty\frac{11(k+1)}{k^3(k+1)k^2} \lt \int_N^\infty\frac{11}{x^5}\,dx=\frac{11}{4N^4}.\]So for example, we get $\zeta(3)\approx1.20190261504$ after $10$ terms, which is pretty close to the next decimal place.