How can we converge to transcendental numbers beyond e?

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e is the limit of an exponential of a number that is approaching one. The exponential makes it want to blow up, but the closeness to 1 keeps that in check. It's really a remarkable number!

My question is, how easy is it to find ways to converge to arbitrary numbers other then e? Almost every number is transcendental so they require such a convergence. A very few of them have fascinatingly concise series, and a remarkable variety of ways to get to that number. But for the rest, how do we get to them?
 
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If f(x) converges to F, then ##e^{f(x)}## converges to ##G=e^F##. For every positive G there is a suitable F=log(G).
This is not limited to exponentiation, this rule works for every continuous function. If f(x) converges to F, then ##2*f(x)## converges to ##G=2*F##.

What do you mean with "get" a number? Here is a transcendal number: ##\pi##.
 
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Every real number can be written as a decimal and, for any irrational number (and most rational numbers), that decimal expansion is an infinite decimal expansion. That is, every such number, which includes all transcendental numbers, can be written as the infinite sequence of it decimal expansions. For example, \pi is the limit of the infinite sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1415926, ...
 
I have a hard time phrasing this question, so my apologies in advance. e and pi are both very special numbers, with very special properties. Not only do they have really cool ways to define them, but a variety of series converge to exactly their values. That is, a variety of series without arbitrary constants. To me, that's awesomely amazing.

The series that let us know the value of e look nothing like the series that let us know the value of pi. This is true in general, isn't it? Two different numbers could be close to each other in value but have wildly different series convergent to them, and this remains true even as they get increasingly close to each other, as long as they remain different. Is that a true statement?
 
Bob3141592 said:
but a variety of series converge to exactly their values

This is true for many series. For example:

\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ... = 2

\frac{1}{1} + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + ... = 2

\frac{3}{1} - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - ... = 2
 
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