$$\sum_{n = 1}^\infty \dfrac 1 {5^{n+1}-5^n+1}$$
I can factorize denominator to $4\times5^n+1$ to confirm the series does not diverge,
But how do I calculate its actual sum?
The series is not a telescoping series nor I can partial factorise. I get confused due to $+1$ in the denominator.
Thanks a lot
$\endgroup$101 Answer
$\begingroup$The series doesn't have a closed form (except for a very complicated one involving Q-Polygamma function, as was said in a comment), however, we can transform it to get much better convergence.
$$\frac{1}{4 \cdot 5^n+1}=\frac{1}{4 \cdot 5^n} \left(1-\frac{1}{4 \cdot 5^n}+\frac{1}{4^2 \cdot 5^{2n}}-\frac{1}{4^3 \cdot 5^{3n}}+\cdots \right)$$
$$\sum_{n=1}^{\infty} \frac{1}{4^k \cdot 5^{k n}}=\frac{1}{4^k} \left( \dfrac{1}{1-\dfrac{1}{5^{k}}}-1 \right)=\frac{1}{4^k (5^k-1)}$$
$$\sum_{n=1}^{\infty} \frac{1}{4 \cdot 5^n+1}=\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{4^k (5^k-1)}=0.06001587909991328$$
Why is this series better? Since it is alternating, it provides upper and lower bounds, unlike the first series, which converges monotonely from below.
It also just gives much better approximations, both in terms of their numerical value and the size of their denominators.
Let's denote:
$$A_N=\sum_{n=1}^{N} \frac{1}{4 \cdot 5^n+1}$$
$$B_N=\sum_{k=1}^{N} (-1)^{k+1} \frac{1}{4^k (5^k-1)}$$
Now compare:
$$A_2=\frac{122}{2121}=0.0575200$$
$$A_3=\frac{7027}{118069}=0.0595161$$
$$B_2=\frac{23}{384}=0.0598958$$
$$B_3=\frac{1429}{23808}=0.0600218$$
$$0.0598958<\sum_{n=1}^{\infty} \frac{1}{4 \cdot 5^n+1}<0.0600218$$
Here is the plot of both $A_N$ and $B_N$ up to $N=10$
