%204
5. From another quite different starting point, {\it 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\footnote{$^{24}$}{%
The following considerable simplification of {\it Euler's\/} method for obtaining
the expansion is due to {\it Schr\"oter\/} (Ableitung der Partialbruch- unde Produkt%-
entwicklungen f\"ur die trigonometrischen Funktionen. Zeitschrift f\"ur Math.~u.~%
Phys., Vol.~13, p.~254. 1868).
}. At the same time, it will give us the radius of convergence of the
series {\bf 115} and {\bf 116} (v.~{\bf 241}).
%205
We have, as was just shewn,
$$\cot y = {1\over 2}\left(\cot{y\over 2}-\tan{y\over 2}\right)$$
or
$$\cot\pi x = {1\over 2}\left\{\cot{\pi x\over 2}+\cot{\pi(x\pm 1)\over 2}\right\},\leqno(*)$$
a formula in which we may, on the right, take either of the signs $\pm$.
Let $x$ be {\it an arbitrary real number distinct from $0$, $\pm 1$, $\pm 2$, $\ldots$,}
whose value will reamain fixed in what follows. Then
$$\pi x\cot\pi x = {\pi x\over 2}\left\{\cot{\pi x\over 2}+\cot{\pi(x+1)\over 2}\right\}$$
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
$$\pi x\cot\pi x = {\pi x\over 4}\left\{\cot{\pi x\over 4}+\left[\cot{\pi(x+1)\over 4}+\cot{\pi(x-1)\over 4}\right]+\cot{\pi(x+2)\over 4}\right\}.$$
A third similar step gives, for $\pi x\cot\pi x$, the value
$${\pi x\over 8}\left\{\matrix{&+\cot{\pi(x+1)\over 8}&+\cot{\pi(x+2)\over 8}&+\cot{\pi(x+3)\over 8}&\cr
\cot{\pi x\over 8}&&\bullet&&+\cot{\pi(x+4)\over 8}\cr
&+\cot{\pi(x-1)\over 8}&+\cot{\pi(x-2)\over 8}&+\cot{\pi(x-3)\over 8}&}\right\},$$
since here each pair of terms which occupy symmetrical positions
relatively to the centre ($\bullet$) of the aggregate in the curly brackets give,
except for a factor ${1\over 2}$, a term of the precdeing aggregate, in accordance
with the formula (*). If we proceed thus through $n$ stages, we obtain
for $n > 1$
$$\pi x\cot\pi x = {\pi x\over2^n}\left\{\cot{\pi x\over 2^n}+\sum_{\nu=1}^{2^{n-1}-1}\left[\cot{\pi(x+\nu)\over 2^n}+\cot{\pi(x-\nu)\over 2^n}\right]-\tan{\pi x\over 2^n}\right\}.\leqno(\dagger)$$
Now by {\bf 115},
$$\lim_{z\to 0} z\cot z = 1$$
and hence for each $\alpha\ne 0$
$$\lim_{n\to\infty} {1\over 2^n}\cot{\alpha\over 2^n} = {1\over\alpha};$$
if in the above expression we let $n\to\infty$ and, {\it at first tentatively,} carry
out the limiting process for each term separately, we obtain the ex%-
pansion
$$\pi x\cot\pi x = 1+x\sum_{\nu=1}^\infty\left({1\over x+\nu}+{1\over x-\nu}\right)-0 = 1+2x^2\sum_{\nu=1}^\infty{1\over x^2-\nu^2}.$$
We proceed to show that this in general {\it faulty mode of passage to the
limit\/} has, however, led in this case to a right result.
We first note that the series converges absolutely for every
$x\ne\pm 1$, $\pm 2$, $\ldots$, by {\bf 70}, 4, since the absolute values of its terms
%206
are asymptotically equal to those of the series $\sum{1\over\nu^2}$. 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 $(\dagger)$ for $\pi x\cot\pi x$, as follows\footnote{$^{25}$}{%
Cf.~Footnote 7, p.~194.
}:
$$\pi x\cot\pi x = {\pi x\over 2^n}\left\{\cot{\pi x\over 2^n}-\tan{\pi x\over 2^n}+\sum_{\nu=1}^k\left[\cdots\right]\right\}+{\pi x\over 2^n}\left\{\sum_{\nu=k+1}^m\left[\cdots\right]\right\}.$$
(In the square brackets we have of course to insert the same expression
as occurs in $(\dagger)$.) 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 {\bf 41}, 9, and
we have
$$\lim_{n\to\infty} A_n = 1+2x^2\sum_{\nu=1}^k{1\over x^2-\nu^2}.$$
Also $B_n$ is preciesely $\pi x\cot\pi x-A_n$, hence $\lim B_n$ certainly exists.
Let $r_k$ denote its value, depending as it does upon the chosen value $k$
thus
$$\lim_{n\to\infty} B_n = r_k = \pi x\cot\pi x-\left[1-2x^2\sum_{\nu=1}^k{1\over x^2-\nu^2}\right].$$
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
$$\cot(a+b)+\cot(a-b) = {-2\cot a\over{\sin^2 b\over\sin^2 a}-1}$$
and hence
$$\cot{\pi(x+\nu)\over 2^n}+\cot{\pi(x-\nu)\over 2^n} = {-2\cot\alpha\over{\sin^2\beta\over\sin^2\alpha}-1}$$
writing for the moment ${\pi x\over 2^n} = \alpha$ and ${\pi\nu\over 2^n} = \beta$, for short.
As $2^n > k > 6|x|$, we certainly have $|a| = \left|\pi x\over 2^n\right| < 1$ and so\footnote{$^{26}$}{%
For the sake of later applications we make these estimates in the above
rough form.
}
$$|\sin\alpha| = \left|\alpha-{\alpha^3\over 3!}+\cdots\right|\le|\alpha|\left(1+{1\over 3!}+{1\over 5!}+\cdots\right) < 2|\alpha|.$$
Since, further $0 < \beta < {\pi\over 2} < 2$, we have\footnote{$^{27}$}{%
Cf.~p.~200.
}
$$\sin\beta = \beta\left(1-{\beta^2\over 2\cdot 3}\right)+{\beta^5\over 5!}\left(1-{\beta^2\over 6\cdot 7}\right)+\cdots > {\beta\over 3}.$$
%207
Hence
$$\left|{\sin\beta\over\sin\alpha}\right| > {\beta\over 6|\alpha|} = {\nu\over 6|x|} > 1,$$
the latter, because $\nu > k > 6|x|$. It therefore follows that (for $\nu > k$)
$$\left|\cot{\pi(x+\nu)\over 2^n}+\cot{\pi(x-\nu)\over 2^n}\right|\le{2\left|\cot{\pi x\over 2^n}\right|\over{\nu^2\over 36\,x^2}-1}$$
and hence
$$|B_n|\le\left|{\pi x\over 2^n}\cot{\pi x\over 2^n}\right|\cdot\sum_{\nu=k+1}^m{72\,x^2\over \nu^2-36\,x^2}.$$
The factor outside the sign of summation is -- quite roughly estimated
-- certainly $< 3$; for $\left|{\pi x\over 2^n}\right|$ was $< 1$, and for $|z| < 1$ we have{$^{26}$}
$$|z\cot z| = \left|{1-{z^2\over 2!}+{z^4\over 4!}-+\cdots\over1-{z^2\over 3!}+{z^4\over 5!}-+\cdots}\right| < {1+{1\over 2!}+{1\over 4!}-+\cdots\over1-{1\over 3!}-{1\over 5!}-+\cdots} < 3.$$
Accordingly,
$$|B_n| < 216\,x^2\cdot\sum_{\nu=k+1}^m{1\over\nu^2-36\,x^2} < 216\,x^2\cdot\sum_{\nu=k+1}^\infty{1\over\nu^2-36\,x^2}.$$
But this is a number quite independent of $n$, so that we may also
write
$$|\lim B_n| = |r_k|\le 216\,x^2\cdot\sum_{\nu=k+1}^\infty{1\over\nu^2-36\,x^2}.$$
But the bound above which we have thus obtained for $r_k$ is equal
to the remainder, after the $k^{\hbox{\sevenrm th}}$ term, of a convergent series\footnote{$^{28}$}{%
The convergence is obtained just as simply as, previously, that of the
series $\sum{1\over x^2-\nu^2}$.
}. Given
$\varepsilon = 0$, we can therefore choose $k_0$ so large that, for every $k > k_0$,
we have
$$|r_k| < \varepsilon.$$
If we refer back to the meaning of $r_k$, we see that this implies
$$\lim_{k\to\infty} \left\{\pi x\cot\pi x-\left[1+2x^2\sum_{n=1}^k{1\over x^2-\nu^2}\right]\right\} = 0,$$
or as asserted,
$$\pi x\cot\pi x = 1+2x^2\sum_{\nu=1}^\infty{1\over x^2-\nu^2},\eqno\hbox{\bf 117.}$$
-- a formula which is thus proved valid for every $x\ne 0$, $\pm 1$, $\pm 2$, $\cdots$.
\bye