I had so much fun deriving that series which converges to pi the other day that I thought it’d be fun to occasionally whip up such series for other irrational numbers. Today I’ll do .

There are many different series that would do the job, the one I’ll derive is found by expanding around as a power series.

Indeed, for satsfying we have

.

This series is guaranteed to converge for all , and to diverge for all and . The number 2 itself sits right on the boundary of the circle of convergence, so although we would like to say

,

we first need to show that the series is convergent. Consider the sequence . This sequence is monotonically decreasing, and we will show that for all . Firstly,

.

We now assume that for some integer and show that . This is true, since

,

where the last inequality holds for . Thus implies , and so by induction for all .

To show convergence of our series, we note we have shown enough to invoke the alternating series test. So it is true that

converges, and by continuity it must equal . Truncating the series at n=10,000,000 gives

which is correct up 11 decimal places.

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