To how many decimals is $\sum_ {k=1}^\infty \frac{k}{\sqrt{k!}} = \frac{49850839\,\pi}{29567947}$ correct? Consider:
$$\sum_ {k=1}^\infty \frac{k}{\sqrt{k!}} = \frac{49850839\,\pi}{29567947}$$
This is, as far as I'm able to check with my software, correct to at least 167 decimals.
If anyone has the resources and time to perform more accurate calculations to determine where LHS and RHS start to differ, it would be very much appreciated.
It would be even better if someone could show it analytically.
 A: 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.
