莉可丽丝二人组联手攻克普特南数学B6题梗图网

B6 Evaluate
$∑{k=1}^{∞} (-1)^{k-1}/k ∑ 1/(k2^n + 1)$}^{∞

Let $S = ∑{k=1}^{∞} S_k$, where $S_k = (-1)^{k-1}/k ∑ 1/(k2^n + 1)$, be the summation in the problem statement.}^{∞
For an odd positive integer $a$, let $T_a = ∑{k=0}^{∞} S$

Lemma: Let $f : \mathbb{N}^2 \to \mathbb{N}$ be defined as $f(x_1,x_2)=(2x_1-1)\cdot2^{x_2-1}$ Then, $f$ is a bijection.

Proof of Lemma:
Note that it is trivial that every $n \in \mathbb{N}$ can be uniquely expressed in binary representation. Since $2x_1 - 1$ is always odd, $f(x_1,x_2)$ contains exactly $x_2 - 1$ powers of 2 in its prime factorization, hence, $f(x_1,x_2)$ is a binary number with $x_2 - 1$ trailing zeroes. Since every positive integer can be uniquely expressed in binary, in particular, every positive integer with $x_2 - 1$ powers of 2 in its prime factorization can be uniquely expressed in the form $(2x_1 - 1) \cdot 2^{x_2-1}$, since $2x_1 - 1$ can obviously take on any arbitrary odd value. Let $E_k$ be the set of positive integers with $k$ powers of 2 in its prime factorization. Since the family of sets ${E_k}_{k\geq0}$ is disjoint (as a given positive integer obviously can't have both $a$ and $b$ powers of 2 in its prime factorization where $a \neq b$) and covers the entirety of $\mathbb{N}$ (as every positive integer has a finite number of powers of 2 in prime factorization), it follows that $f$ is bijective.

Observe that from the lemma and the fact that $S$ is absolutely convergent (which can be derived from a bounding argument using the fact that $|S_k| = \frac{1}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n +1} < \frac{1}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n} = \frac{2}{k^2}$ hence $\sum_{k=1}^{\infty}|S_k| < 2\sum_{k=1}^{\infty}\frac{1}{k^2}$, the RHS which is known to converge to $\frac{\pi^2}{3}$, hence the LHS is convergent) we can express $S$ in the alternate equivalent form $S = \sum_{b=1}^{\infty} T_{2b-1}$.

We have
$T_{2b-1} = \sum_{k=0}^{\infty} S_{(2b-1)\cdot2^k} = \sum_{k=0}^{\infty} \frac{(-1)^{(2b-1)\cdot2^k -1}}{(2b-1)\cdot2^k} \sum_{n=0}^{\infty} \frac{1}{(2b-1)\cdot2^k \cdot 2^n +1}$.
Since the exponent $(2b-1)\cdot2^k -1$ is even only for $k=0$ and odd for all $k>0$, we can rewrite this as
$T_{2b-1} = \frac{1}{2b-1}\sum_{n=0}^{\infty}\frac{1}{(2b-1)2^n +1} - \sum_{k=1}^{\infty}\frac{1}{(2b-1)\cdot2^k}\sum_{n=0}^{\infty}\frac{1}{(2b-1)\cdot2^k \cdot 2^n +1}$.

Rearranging the terms of the above summation using the distributive property, we obtain
$T_{2b-1} = \sum_{k=0}^{\infty} \left( \frac{1}{2b-1} - \sum_{n=1}^{k}\frac{1}{(2b-1)2^n} \right) \frac{1}{(2b-1)2^k +1}$.

Note that by the finite geometric series formula,
$\sum_{n=1}^{k}\frac{1}{(2b-1)2^n} = \frac{1}{2b-1} - \frac{1}{(2b-1)2^k}$. Hence, for each
$k$ the coefficient $\frac{1}{2b-1} - \sum_{n=1}^{k}\frac{1}{(2b-1)2^n}$ equals
$\frac{1}{(2b-1)2^k}$ and we can simplify the whole expression for
$T_{2b-1}$ to $T_{2b-1} = \sum_{k=0}^{\infty} \frac{1}{(2b-1)2^k} \cdot \frac{1}{(2b-1)2^k +1}$.

Since we established from the Lemma that $S = \sum_{b=1}^{\infty} T_{2b-1}$,
we can rewrite $S$ as $S = \sum_{k=1}^{\infty} \frac{1}{k} \cdot \frac{1}{k+1}$, which is a well-known telescoping sum that converges to $S = \boldsymbol{1}$.

底部文字：WHOLESOME ε>0

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