Evaluate 1/Gamma(3/5) times the sum from n = 1 to infinity of (1/10)^(8n/5 + 1) divided by Gamma(8n/5 + 1).

Step-by-step solution

1. 1

   Recognise the shape of the series. The summand $\dfrac{z^{\alpha n+1}}{\Gamma(\alpha n+1)}$ with $z=\tfrac{1}{10}$ and $\alpha=\tfrac85$ is a Mittag-Leffler-type term. There is no elementary closed form, so the practical route is to sum numerically — and because $z<1$ while $\Gamma(\alpha n+1)$ grows super-exponentially, only a handful of terms carry any weight.

2. 2

   Compute the $n=1$ term. The exponent is $\tfrac85(1)+1=2.6$, so the term is $\dfrac{10^{-2.6}}{\Gamma(2.6)}=\dfrac{2.511886\times10^{-3}}{1.4296246}=1.757025\times10^{-3}$.

3. 3

   Compute the $n=2$ and $n=3$ terms. For $n=2$ the exponent is $4.2$: $\dfrac{10^{-4.2}}{\Gamma(4.2)}=\dfrac{6.309573\times10^{-5}}{7.7566895}=8.134364\times10^{-6}$. For $n=3$ the exponent is $5.8$: $\dfrac{10^{-5.8}}{\Gamma(5.8)}=\dfrac{1.584893\times10^{-6}}{85.621738}=1.851041\times10^{-8}$.

4. 4

   Check that the tail is negligible. The $n=4$ term is $2.58\times10^{-11}$ and the $n=5$ term is $2.48\times10^{-14}$; each successive term drops by roughly two to three orders of magnitude. Truncating after $n=3$ therefore fixes the sum to well beyond six significant figures: $\sum_{n=1}^{\infty}\frac{10^{-(1.6n+1)}}{\Gamma(1.6n+1)}=1.765178\times10^{-3}.$

5. 5

   Divide by $\Gamma(3/5)$. With $\Gamma(0.6)=1.4891922$, $\frac{1.765178\times10^{-3}}{1.4891922}=1.185326\times10^{-3}.$

6. 6

   State the value. The whole expression equals $0.00118533$ to six significant figures; the $n=1$ term alone already supplies 99.5 percent of it.