To how many decimals is $\sum_ {k=1}^\infty \frac{k}{\sqrt{k!}} = \frac{49850839\,\pi}{29567947}$ correct?

Solution 1:

For $k>n+1>1$, we have $$ \frac k{\sqrt{k!}}< \frac{\sqrt 2\,\sqrt{k(k-1)}}{\sqrt{k!}}=\sqrt{\frac{2}{(k-2)!}}\le\sqrt{\frac{2}{n!n^{k-n-2}}}=\sqrt{\frac{2}{n!}}\cdot \sqrt{\frac1n}^{k-n-2}$$ so that $$ \sum_{k=n+2}^\infty\frac k{\sqrt{k!}}<\sqrt{\frac{2}{n!}}\sum_{d=0}^\infty\sqrt{\frac1n}^d=\frac{\sqrt{\frac{2}{n!}}}{1-\sqrt{\frac1n}}$$ Therefore, the finite sum $$ \sum_{k=0}^{37}\frac k{\sqrt{k!}}$$ differs from the true value by less than $\frac65{\sqrt{\frac 2{36!}}}{}<3\cdot 10^{-21}$. Computing said finite sum with enough precision to have an error $<10^{-22}$ in each summand therefore gives us an error $<67\cdot 10 ^{-22}<10^{-20}$. A quick numerical computation (using PARI/GP,for example) therefore tells us that all digits in $$ \sum_{k=0}^\infty\frac k{\sqrt{k!}}\approx 5.296648752031635\color{red}7055$$ are correct. On the other hand, $$\frac{49850839\pi}{29567947}= 5.296648752031635\color{red}973031534\ldots$$

In doesn't take many summands to compute the series to higher precision (1000 digits, say) and obtain the continued fraction expansion of the result (or the result divided by $\pi$) as far as the precision allows. Neither of these suggests that the series, or the series divided by $\pi$, is rational.