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$$\sum_{n = 1}^\infty \dfrac 1 {5^{n+1}-5^n+1}$$

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

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    $\begingroup$ You can have a very good (and simple) upper bound using the fact that $4\times5^n+1>4\times5^n$ $\endgroup$ – Claude Leibovici Feb 2 '16 at 9:20
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    $\begingroup$ If I can approximate the sum it would be lot easier. But what about its actual value? $\endgroup$ – zcahfg2 Feb 2 '16 at 9:23
  • $\begingroup$ The sum $\sum_{k=1}^\infty \frac{1}{1-a^k}$ can be expressed in terms of the Q-Polygamma function mathworld.wolfram.com/q-PolygammaFunction.html Just rewrite $5^{n+1}-5^n=4\times 5^n$ and massage the sum a bit, perhaps? $\endgroup$ – Pierpaolo Vivo Feb 2 '16 at 9:24
  • $\begingroup$ This gives a quite good approximation. The exact value looks diificult (at least to me). $\endgroup$ – Claude Leibovici Feb 2 '16 at 9:24
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    $\begingroup$ Probably the fly is too strong to kill with a simple swatter. We'll need quite a large hammer for this $\endgroup$ – vrugtehagel Feb 2 '16 at 9:27
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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 $N=10$

enter image description here

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