5. From another quite different starting point, Euler obtained an interesting expansion for the cotangent which we proceed to deduce, especially as it is of great importance for many problems in series^{24}. At the same time, it will give us the radius of convergence of the series 115 and 116 (v. 241).

We have, as was just shewn,

cot y = \frac{1}{2} (cot \frac{y}{2} − tan \frac{y}{2} )

or

(∗)  cot πx = \frac{1}{2} \{cot \frac{πx}{2} + cot \frac{π(x ± 1)}{2}\} ,

a formula in which we may, on the right, take either of the signs ±. Let x be an arbitrary real number distinct from 0, ±1, ±2, …, whose value will reamain fixed in what follows. Then

πx cot πx = \frac{πx}{2} \{cot \frac{πx}{2} + cot \frac{π(x + 1)}{2}\}

and applying the formula (*) once more to both functions on the right had side, taking for the first the + and for the second, the − sign, we obtain

πx cot πx = \frac{πx}{4} \{cot \frac{πx}{4} + [cot \frac{π(x + 1)}{4} + cot \frac{π(x − 1)}{4}] + cot \frac{π(x + 2)}{4}\} .

A third similar step gives, for πx cot πx, the value

\frac{πx}{8} \{ \begin{matrix}& + cot \frac{π(x+1)}{8} & + cot \frac{π(x+2)}{8} & + cot \frac{π(x+3)}{8} \\ cot \frac{πx}{8} & & • & & + cot \frac{π(x+4)}{8} \\ & + cot \frac{π(x−1)}{8} & + cot \frac{π(x−2)}{8} & + cot \frac{π(x−3)}{8} \end{matrix} \} ,

since here each pair of terms which occupy symmetrical positions relatively to the centre (•) of the aggregate in the curly brackets give, except for a factor \frac{1}{2}, a term of the precdeing aggregate, in accordance with the formula (*). If we proceed thus through n stages, we obtain for n > 1

(†)  πx cot πx = \frac{πx}{2^{n}} \{cot \frac{πx}{2^{n}} + ∑_{ν=1}^{2^{n−1}−1} [cot \frac{π(x + ν)}{2^{n}} + cot \frac{π(x − ν)}{2^{n}}] − tan \frac{πx}{2^{n}} \} .

Now by 115,

lim_{z→0} z cot z = 1

and hence for each α ≠ 0

lim_{n→∞} \frac{1}{2^{n}} cot \frac{α}{2^{n}} = \frac{1}{α};

if in the above expression we let n → ∞ and, at first tentatively, carry out the limiting process for each term separately, we obtain the expansion

πx cot πx = 1 + x ∑_{ν=1}^{∞} (\frac{1}{x + ν} + \frac{1}{x − ν} ) − 0 = 1 + 2x^{2} ∑_{ν=1}^{∞} \frac{1}{x^{2} − ν^{2}} .

We proceed to show that this in general faulty mode of passage to the limit has, however, led in this case to a right result.

24 The following considerable simplification of Euler’s method for obtaining the expansion is due to Schröter (Ableitung der Partialbruch- unde Produktentwicklungen für die trigonometrischen Funktionen. Zeitschrift für Math. u. Phys., Vol. 13, p. 254. 1868).

We first note that the series converges absolutely for every x ≠ ±1, ±2, …, by 70, 4, since the absolute values of its terms are asymptotically equal to those of the series ∑ \frac{1}{ν^{2}} . Now choose an arbitary integer k > 6|x|, to be kept provisionally fixed. If n is so large that the number 2^{n−1} − 1, which we will denote for short by m, is > k, we then split up the expression (†) for πx cot πx, as follows^{25}:

πx cot πx = \frac{πx}{2^{n}} \{cot \frac{πx}{2^{n}} − tan \frac{πx}{2^{n}} + ∑_{ν=1}^{k} [⋯]\} + \frac{πx}{2^{n}} \{ ∑_{ν=k+1}^{m} [⋯]\} .

(In the square brackets we have of course to insert the same expression as occurs in (†).) The two parts of this expression we denote by A_{n} and B_{n}. Since A_{n} consists of a finite number of terms, the passage to the limit term by term is certainly allowed there, by 41, 9, and we have

lim_{n→∞} A_{n} = 1 + 2x^{2} ∑_{ν=1}^{k} \frac{1}{x^{2} − ν^{2}} .

Also B_{n} is preciesely πx cot πx − A_{n}, hence lim B_{n} certainly exists. Let r_{k} denote its value, depending as it does upon the chosen value k thus

lim_{n→∞} B_{n} = r_{k} = πx cot πx − [1 − 2x^{2} ∑_{ν=1}^{k} \frac{1}{x^{2} − ν^{2}} ] .

Bounds above for the numbers B_{n}, hence for their limit r_{k} and so finally for the difference on the right hand side, may now quite easily be estimated:

We have

cot(a + b) + cot(a − b) = \frac{−2 cot a}{\frac{sin^{2} b}{sin^{2} a} − 1}

and hence

cot \frac{π(x + ν)}{2^{n}} + cot \frac{π(x − ν)}{2^{n}} = \frac{−2 cot α}{\frac{sin^{2} β}{sin^{2} α} − 1}

writing for the moment \frac{πx}{2^{n}} = α and \frac{πν}{2^{n}} = β, for short.

As 2^{n} > k > 6|x|, we certainly have |a| = |\frac{πx}{2^{n}} | < 1 and so^{26}

| sin α| = |α − \frac{α^{3}}{3!} + ⋯| ≤ |α| (1 + \frac{1}{3!} + \frac{1}{5!} + ⋯) < 2|α|.

Since, further 0 < β < \frac{π}{2} < 2, we have^{27}

sin β = β (1 − \frac{β^{2}}{2 · 3}) + \frac{β^{5}}{5!} (1 − \frac{β^{2}}{6 · 7}) + ⋯ > \frac{β}{3} .

Hence

|\frac{sin β}{sin α}| > \frac{β}{6|α|} = \frac{ν}{6|x|} > 1,

the latter, because ν > k > 6|x|. It therefore follows that (for ν > k)

|cot \frac{π(x + ν)}{2^{n}} + cot \frac{π(x − ν)}{2^{n}} | ≤ \frac{2 |cot \frac{πx}{2^{n}} |}{\frac{ν^{2}}{36 x^{2}} − 1}

25 Cf. Footnote 7, p. 194. 26 For the sake of later applications we make these estimates in the above rough form. 27 Cf. p. 200. and hence

|B_{n}| ≤ |\frac{πx}{2^{n}} cot \frac{πx}{2^{n}} | · ∑_{ν=k+1}^{m} \frac{72 x^{2}}{ν^{2} − 36 x^{2}} .

The factor outside the sign of summation is – quite roughly estimated – certainly < 3; for |\frac{πx}{2^{n}} | was < 1, and for |z| < 1 we have^{26}

|z cot z| = |\frac{1 − \frac{z^{2}}{2!} + \frac{z^{4}}{4!} − + ⋯}{1 − \frac{z^{2}}{3!} + \frac{z^{4}}{5!} − + ⋯}| < \frac{1 + \frac{1}{2!} + \frac{1}{4!} − + ⋯}{1 − \frac{1}{3!} − \frac{1}{5!} − + ⋯} < 3.

Accordingly,

|B_{n}| < 216 x^{2} · ∑_{ν=k+1}^{m} \frac{1}{ν^{2} − 36 x^{2}} < 216 x^{2} · ∑_{ν=k+1}^{∞} \frac{1}{ν^{2} − 36 x^{2}} .

But this is a number quite independent of n, so that we may also write

| lim B_{n}| = |r_{k}| ≤ 216 x^{2} · ∑_{ν=k+1}^{∞} \frac{1}{ν^{2} − 36 x^{2}} .

But the bound above which we have thus obtained for r_{k} is equal to the remainder, after the k^{th} term, of a convergent series^{28}. Given ε = 0, we can therefore choose k_{0} so large that, for every k > k_{0}, we have

|r_{k}| < ε.

If we refer back to the meaning of r_{k}, we see that this implies

lim_{k→∞} \{πx cot πx − [1 + 2x^{2} ∑_{n=1}^{k} \frac{1}{x^{2} − ν^{2}} ]\} = 0,

or as asserted,

πx cot πx = 1 + 2x^{2} ∑_{ν=1}^{∞} \frac{1}{x^{2} − ν^{2}} ,  117.

– a formula which is thus proved valid for every x ≠ 0, ±1, ±2, · · ·.

28 The convergence is obtained just as simply as, previously, that of the series ∑ \frac{1}{x^{2}−ν^{2}} .