Alan

The value of 1.5^31515999 exceeds the range of the calculator here, so it just returns ∞ for this.  Interestingly, instead of just returning ∞ for the whole calculation, it returns the result as a product of the result of (1000+1000/31536000) and infinity!! 

$$\left({\mathtt{1\,000}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1\,000}}}{{\mathtt{31\,536\,000}}}}\right){\mathtt{\,\times\,}}{{\mathtt{1.5}}}^{{\mathtt{31\,535\,999}}} = {\mathtt{1\,000.000\: \!031\: \!709\: \!791\: \!983\: \!8}}{\mathtt{\,\times\,}}\left(\infty\right)$$

Basically, you just have to interpret this as a number too big for the calculator here to compute!

I bought Petr Beckmann's "A History of Pi" on my first visit to the US (a large number of years ago!) and still have it.  A fascinating book, not least because of Beckmann's writing style!  It still seems to be available on Amazon.

You are right Melody.  I really must stop doing stuff late at night! 

Impressive calculations, but the quadrilateral isn't unique - see illustration below (drawn to scale in Geogebra). Compare the blue and red quadrilaterals, both on the same black baseline.

2991

$${\mathtt{2\,991}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,992}} = {\mathtt{4\,983}}$$

Start by letting the number be 1000a+100b+10c+d so its reverse is 1000d+100c+10b+a.

Adding these together we get 1001(a+d)+110(b+c).  This must equal 4983.

I'll leave you to take the reasoning further! 

No, this sum has a finite limit:

$$\sum_{k=2}^{\inf}(\frac{-3}{7})^k=\frac{9}{70}$$

See the following figure;

 In this respect it is more like the infinite series 1 + 1/2 + 1/4 + 1/8 etc. which has the finite sum of 2. In other words it is a geometric series, the sum to infinity of which is given by:

sum = a0/(1-r) where a0 is the first term [(-3/7)² here] and r [= (-3/7)] is the ratio between successive terms.  Because r is less than 1, successive terms get smaller, r^∞→0, and the sum converges to a finite value.

$${\mathtt{sum}} = {\frac{{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)\right)}} \Rightarrow {\mathtt{sum}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

$${\frac{{\mathtt{9}}}{{\mathtt{70}}}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

Just noticed this is a very old unanswered question!!

Ah! tan(x)≥sin(x)

Edit.

Melody is correct below!  The correct result is given when tan(x)-sin(x)>=0. i.e. when 0≤x<pi/2 and pi≤x<3pi/2 

Use the calculator on the Home page here.

The slope is (difference in y_values/difference in x-values).

$${\mathtt{slope}} = {\frac{\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}} \Rightarrow {\mathtt{slope}} = -{\mathtt{1.5}}$$

Are you sure the question is complete?  The graph below shows that all values in the region shaded in red (which extends up to infinity!) satisfies the condition ≥sin(x).

If tg(x) is restricted to the range 0 to 2pi as well, then it's all the red values that are shown above zero, stretching up to 2pi.