Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$

Proof. The function
$$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$
is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that
$$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$
If we make change of variables in the latter integral $t\to 1-t$ we obtain
$$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$
On the other hand, using integration by parts, we obtain
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$
Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$
Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain
$$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$
On the other hand  $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t+\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=\int_{0}^1\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{6}$$
Substracting the last two equalities we obtain  $$\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{12}-\frac{1}{2}\log\left(\frac{1}{2}\right)^2$$
and thus $\displaystyle S\left(\frac{1}{2}\right)=\frac{\pi^2}{12}+\frac{1}{2}\log\left(\frac{1}{2}\right)^2$.

Comments

  1. Unknown11 February 2020 at 06:39

    Редно е да се отбележи, че тази задача (поне за България) е поставена от проф. Димитър Скордев преди повече от 50 години. Виж Иван Проданов, Н. Хаджииванов, Иван Чобанов, Сборник от задачи по ДИС (функции на една променлива), Второ стереотипно издания, УИ "Св. Кл. Охридски", София, 1992.
    Прекрасно е човек сам да се справи с такава задача, но все пак трябва да се цитира и сборника (в който подробно е изложено решението).

    ReplyDelete
    Replies
    1. Saposto11 February 2020 at 13:27

      Разбира се, съжалявам за пропуска и благодаря за уточнението! Вече е отбелязано.

      Delete

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