9
$\begingroup$

Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms

My Attempt:

The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$

So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$

$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$ which is obviously absurd since all terms of $S$ are positive.

But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$

What is wrong with the previous approach

$\endgroup$
1
  • 9
    $\begingroup$ $$\int_0^1\dfrac{\textrm dx}{1-x^2}\textrm{ diverges.}$$ $\endgroup$ Aug 25, 2021 at 16:27

6 Answers 6

5
$\begingroup$

As was mentioned in the comments, the integral $$\int_0^1\frac{dx}{1-x^2}$$ diverges, so you cannot split the integral $$\int_0^1\left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)dx$$ up into parts.

$\endgroup$
4
$\begingroup$

Another similar approach: consider the Beta integral $$\int_0^1 (1-x)^2 x^n d x = B(3, n+1) = \frac{\Gamma(3)\Gamma(n+1)}{\Gamma(3 + n+1)} = \frac{2! \cdot n!}{(n+3)!} = \frac{2}{(n+1)(n+2)(n+3)}$$

We get $$\sum_{n\ge 0} \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac{1}{2} \int_{0}^1 (1-x)^2 (\sum_{n\ge 0}x^{4n}) dx=\\ = \frac{1}{2} \int_0^1 \frac{(1-x)^2}{1 - x^4} d x = \frac{\log 2}{4}$$

$\endgroup$
1
  • $\begingroup$ this does not answer the question $\endgroup$
    – clathratus
    Aug 25, 2021 at 17:42
4
$\begingroup$

Try this. $$ F(x) := \sum_{i=0}^\infty\frac{x^{4i+3}}{(4i+1)(4i+2)(4i+3)},\quad |x| \le 1 $$ Problem: compute $F(1)$.
Differentiate, to get geometric series $$ F'''(x) = \sum_{i=0}^\infty x^{4i} = \frac{1}{1-x^4},\quad |x|<1 $$ Now integrate this three times, using $F(0) = F'(0) = F''(0)=0$. $$ F(x) = \frac{x^2-1}{4}\arctan x + \frac{x^2}{4}\operatorname{atanh} x +\frac{1}{8}\log(1+x) - \frac{1}{8}\log(1-x) +\frac{x}{4}\log(1-x^2)-\frac{x}{4}\log(1+x^2) $$ and then take limit $$ \lim_{x\to 1^-}F(x) = \frac{\log 2}{4} $$

$\endgroup$
2
$\begingroup$

We have $$ \begin{array}{l} S = \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{5 \cdot 6 \cdot 7}} + \frac{1}{{9 \cdot 10 \cdot 11}} + \cdots = \\ = \sum\limits_{0 \le k} {\frac{1}{{\left( {4k + 1} \right)^{\,\overline {\;3\,} } }}} = \sum\limits_{0 \le k} {\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}}} = \sum\limits_{0 \le k} {t_{\,k} } \\ \end{array} $$ where $ x^{\,\overline {\,k\,} } $ represents the Rising Factorial

Then we have $$ \begin{array}{l} t_{\,0} = \frac{1}{{1 \cdot 2 \cdot 3}} = \frac{1}{6} \\ \frac{{t_{\,k + 1} }}{{t_{\,k} }} = \frac{{\Gamma \left( {4k + 5} \right)}} {{\Gamma \left( {4k + 8} \right)}}\frac{{\Gamma \left( {4k + 4} \right)}}{{\Gamma \left( {4k + 1} \right)}} = \frac{{\left( {4k + 1} \right)^{\,\overline {\;4\,} } }}{{\left( {4k + 4} \right)^{\,\overline {\;4\,} } }} = \prod\limits_{j = 0}^3 {\frac{{\left( {4k + 1 + j} \right)}}{{\left( {4k + 4 + j} \right)}}} = \\ = \prod\limits_{j = 0}^3 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 1 + j/4} \right)}}} = \prod\limits_{j = 0}^2 {\frac{{\left( {k + 1/4 + j/4} \right)}}{{\left( {k + 5/4 + j/4} \right)}}} \\ \end{array} $$ that is the ratio of the consecutive addenda is a rational function of $k$.

So we can represent the sum as the Generalized Hypergeometric function $$ S = \frac{1}{6}{}_4F_{\,3} \left( {\left. {\begin{array}{*{20}c} {1,\frac{1}{4},\;\frac{2}{4},\;\frac{3}{4}\;} \\ {1 + \frac{1}{4},\;1 + \frac{2}{4},\;1 + \frac{3}{4}} \\ \end{array}\;} \right|\;1} \right) $$

which gives $S=0.1723286 \ldots$

$\endgroup$
0
2
$\begingroup$

Thinking about generalized harmonic numbers $$\frac{1}{(4 i+1) (4 i+2) (4 i+3)}=\frac{1}{2 (4 i+1)}+\frac{1}{2 (4 i+3)}-\frac{1}{(4i+2)}$$ $$S_n=\sum_{i=0}^n\frac{1}{(4 i+1) (4 i+2) (4 i+3)}$$ $$\sum_{i=0}^n \frac{1}{ (4 i+1)}=\frac{1}{4} \left(\psi \left(n+\frac{5}{4}\right)-\psi \left(\frac{1}{4}\right)\right)$$ $$\sum_{i=0}^n \frac{1}{ (4 i+3)}=\frac{1}{4} \left(\psi \left(n+\frac{7}{4}\right)-\psi \left(\frac{3}{4}\right)\right)$$ $$\sum_{i=0}^n \frac{1}{ (4 i+2)}=\frac{1}{4} \left(\psi \left(n+\frac{3}{2}\right)-\psi \left(\frac{1}{2}\right)\right)$$ $$S_n=\frac{1}{4} \left(H_{2 n+\frac{3}{2}}-H_{n+\frac{1}{2}}\right)$$ Using the asymptotics, then $$S_n=\frac{\log (2)}{4}-\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right)$$

Thinking about the gaussian hypergeometric function $$\sum_{i=0}^\infty\frac{x^{2i}}{(4 i+1) (4 i+2) (4 i+3)}=\frac{3 x \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};x^2\right)+x \, _2F_1\left(\frac{3}{4},1;\frac{7}{4};x^2\right)-3 \tanh ^{-1}(x)}{6 x}$$ and, if $x \to 1$, the limit.

Just, for the fun, notice that for $x=0$, the result is already $\frac 16=0.1667$ not "so far" from $\frac{\log(2)}4=0.1733$.

$\endgroup$
2
$\begingroup$

Find the limit

$$\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$$

The lim can be writen as:

$$S=\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{(4k+1)(4k+2)(4h+3)}$$

$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{4k+2}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\sum_{k=0}^{\infty}\left(\frac{1}{(4k+1)(4k+2)}-\frac{1}{(4k+3)(4k+3)}\right)$$

$$=\frac{1}{2}\sum_{k=0}^{\infty}\left\{\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)\right\}=\frac{1}{2}(S_1-S_2)$$

Now we will use the special function (digamma function):

$$\Psi(x)=\frac{d}{dx}\ln\Gamma(x)$$

and the identity

$$\sum_{k=0}^{\infty}\left(\frac{1}{k+y}-\frac{1}{k+x}\right)=\Psi(x)-\Psi(y)$$

With $y=\frac{1}{4},\quad x=\frac{2}{4}$:

$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+1}-\frac{4}{2k+2}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{2k+2}\right)=\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)\rightarrow$$

$$S_1=\frac{1}{4}\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)$$

With $y=\frac{2}{4},\quad x=\frac{3}{4}$:

$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+2}-\frac{4}{2k+3}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+2}-\frac{1}{2k+3}\right)=\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)\rightarrow$$

$$S_2=\frac{1}{4}\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)$$

Hence

$$S=-\frac{1}{8}\left(\Psi\left(\frac{1}{4}\right)-2\Psi\left(\frac{1}{2}\right)+\Psi\left(\frac{3}{4}\right)\right)$$

The particular values of $\Psi(x)$ are:

$\psi\left(\frac{1}{2}\right)=-C-2\ln2$

$\psi\left(\frac{1}{4}\right)=-C-\frac{\pi}{2}-3\ln2$

$\psi\left(\frac{3}{4}\right)=-C+\frac{\pi}{2}-3\ln2$

Whwre $C$ is the Euler's constante.

Replacing in $S$ we obtine

$$S=\frac{\ln2}{4}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .