# Chaitanya's Random Pages

## July 24, 2011

### Tannery’s theorem

Filed under: mathematics — ckrao @ 12:19 pm

Here is a result that I feel deserves to be better known.

Tannery’s Theorem (series form): Suppose that

1. $\displaystyle s(n) = \sum_{k=1}^{\infty} f_k(n)$ is a convergent sum for all $n \geq 1$.
2. $\displaystyle \lim_{n \rightarrow \infty} f_k(n)$ exists and is equal to $f_k$ for all $k \geq 1$.
3. $\displaystyle |f_k(n)| \leq M_k$ for all $k \geq 1, n \geq 1$.
4. $\sum_{k=1}^{\infty} M_k$ converges.

Then $\displaystyle \sum_{k=1}^{\infty} f_k$ converges and is equal to $\lim_{n\rightarrow \infty} s(n)$. In other words, both sides of the following are well-defined and equal:

$\displaystyle \lim_{n\rightarrow \infty} \sum_{k=1}^{\infty} f_k(n) = \sum_{k=1}^{\infty} \lim_{n\rightarrow \infty} f_k(n)$

Note the similarity with the Weierstrass M-test which states that if $f_k(x) \leq M_k$ for all k and if $\sum_{k=1}^{\infty} M_k$ converges, then $\sum_{k=1}^{\infty} f_k(x)$ converges uniformly to $s(x) = \sum_{k=1}^{\infty} f_k(x)$. If each $f_k$ is continuous, so is s. Tannery’s theorem follows from this by the continuity of $f_k$ and s at $\infty$ [1]. It can be proved by standard $\epsilon-$ type arguments.

Tannery’s theorem can also be considered a consequence of Lebesgue’s dominant convergence theorem applied to the sequence space $\ell^1$ with counting measure (hence integrals become sums). It gives a condition (namely, dominance by a convergent series) under which the order of limits and infinite sums may be interchanged.

One common application of this result in showing that the following two expressions for exp(x) are equal:

$\displaystyle \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n = \sum_{k = 0}^{\infty} \frac{x^k}{k!}.$

Setting $f_k(n) = \binom{n}{k}\frac{x^k}{n^k}$ it can be shown that $f_k = \lim_{n \rightarrow \infty} f_k(n) = \frac{x^k}{k!}$ and Tannery’s theorem can be applied with the upper bound $M_k = \frac{|x|^k}{k!}$.

Tannery’s theorem can also exist in product form:

Tannery’s Theorem (product form): Suppose that

1. $\displaystyle p(n) = \prod_{k=1}^{\infty} (1 + f_k(n))$ is a convergent product for all $n \geq 1$.
2. $\displaystyle \lim_{n \rightarrow \infty} f_k(n)$ exists and is equal to $f_k$ for all $k \geq 1$.
3. $\displaystyle |f_k(n)| \leq M_k$ for all $k \geq 1, n \geq 1$.
4. $\sum_{k=1}^{\infty} M_k$ converges.

Then $\displaystyle \prod_{k=1}^{\infty} f_k$ converges and is equal to $\lim_{n\rightarrow \infty} p(n)$. In other words, both sides of the following are well-defined and equal:

$\displaystyle \lim_{n\rightarrow \infty} \prod_{k=1}^{\infty} (1 + f_k(n)) = \prod_{k=1}^{\infty} \lim_{n\rightarrow \infty} (1 + f_k(n))$

This follows from the fact that $\prod(1 + a_n)$ converges absolutely if and only if $\sum a_n$ converges absolutely (taking logarithms helps to see this).

One application of this result is in proving Euler’s infinite product formula for the sine function:

$\displaystyle \sin \pi z = \pi z \prod_{k=1}^{\infty} \left(1-\frac{z^2}{k^2}\right)$

To prove this (ref: [2]), firstly, it can be shown that for any complex z,

$\displaystyle \sin z = \lim_{n\rightarrow\infty} F_n(z),$

where $n = 2m+1$ is odd and

$\displaystyle F_n(z) := \frac{1}{2i} \left \{\left(1 + \frac{iz}{n}\right)^2 - \left(1 - \frac{iz}{n}\right)^n\right \} = z\prod_{k=1}^m \left(1 - \frac{z^2}{n^2 \tan^2 (k \pi/n)}\right).$

[In brief, set z to $n \tan \theta$ and show $F_n(n\tan \theta) = \sec^n \theta \sin(n\theta)$.  Hence the 2m+1 zeros of $F_n$ are at $n\tan \left(\frac{k\pi}{2m+1}\right)$. Finally construct the polynomial with these numbers as roots.]

Applying the product form of Tannery’s theorem with $f_k(n) = - \frac{z^2}{n^2\tan^2 (k\pi/n)}$ and $M_k = \frac{|z|^2}{k^2\pi^2}$ gives

$\displaystyle \sin z = z \lim_{m \rightarrow \infty} \prod_{k=1}^{\infty} (1 + f_k(m)) = z \prod_{k=1}^{\infty} \lim_{m \rightarrow \infty} (1 + f_k(m)) = z\prod_{k=1}^{\infty}\left(1 - \frac{z^2}{k^2\pi^2}\right).$

Finally, replace z with $\pi z$ for the desired result. In my next mathematical post I plan to show some consequences of this beautiful formula for sine.

#### References

[1] J. Hofbauer, A Simple Proof of 1 + 1/2^2 + 1/3^2 + ⋯ = π^2/6 and Related Identities, The American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 196-200

[2] P. Loya, Amazing and Aesthetic Aspects of Analysis: On the incredible infinite, available at http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdf