4
$\begingroup$

Let $$a_n=\frac{1}{n+\frac{1}{2}}\left [\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right]^2,$$ and $$b_n=\frac{1}{n^2}.$$ On the ground of hydrogen atom quantum physics, it was shown in http://arxiv.org/abs/1510.07813 (Quantum Mechanical Derivation of the Wallis Formula for $\pi$, by T. Friedmann and C. R. Hagen) that $$\lim\limits_{n\to\infty}\frac{a_n}{b_n}=1.$$ Therefore, since $$\sum\limits_{n=1}^\infty b_n=\frac{\pi^2}{6},$$ the series $\sum\limits_{n=1}^\infty a_n$ is convergent, according to the limit comparison test. Alternatively we can use direct comparison test, because $a_n\le b_n$ for all $n$ (also from physics).

Can the sum $\sum\limits_{n=1}^\infty a_n$ be calculated explicitly?

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ Mathematica Programming Lab ( lab.wolframcloud.com ) gives $4(\pi-2)/\pi$. $\endgroup$
    – Johannes Trost
    Commented Aug 1, 2016 at 12:08
  • 1
    $\begingroup$ Just by-the-way, $\Gamma(s)/\Gamma(s+a)\sim s^{-a}$ for fixed $a$, by (so-called) Watson's Lemma, so this asymptotic is even easier than invocation of Laplace-Stirling. So $a_n\ll 1/n^2$, giving convergence directly. $\endgroup$
    – paul garrett
    Commented Aug 1, 2016 at 13:07
  • $\begingroup$ Thanks! However I think Watson's Lemma gives $a_n\sim\frac{1}{(n+1/2)n}$. So $a_n<b_n$ (but not $a_n\ll b_n$) which is enough for convergence. $\endgroup$
    – Zurab Silagadze
    Commented Aug 1, 2016 at 13:18
  • $\begingroup$ Ah, sorry, by $\ll$ I meant "is less than a positive constant multiple of"... $\endgroup$
    – paul garrett
    Commented Aug 1, 2016 at 13:30

5 Answers 5

Reset to default
5
$\begingroup$

Using Christian Krattenthaler's hyp.m tells you the following - the important information is at the end, and the result agrees with Johannes' comment.

In[1]:= <<hyp.m

Out[1]= ▒

In[10]:= S = SUM[1/(n+1/2)*(Gamma[n]^2/Gamma[n+1/2]^2),{n,1,Infinity}]

         Infinity
          -----                2
           \           Gamma[n]
Out[10]=    >    ---------------------
           /      1            1     2
          -----  (- + n) Gamma[- + n]
           n=1    2            2

In[19]:= SF = S/.SUMF

              [ 1, 1, 1    ]
              |            |      2
         2  F |  5  3  ; 1 | Ga(1)
           3 2|  -, -      |
              [  2  2      ]
Out[19]= --------------------------
                       3 2
                  3 Ga(-)
                       2

In[21]:= SF/.SListe

Be sure to apply "FOrdne" before using the following information!

                        2
           2 S3261 Ga(1)
Out[21]= {{--------------}}
                   3 2
              3 Ga(-)
                   2

In[22]:= SF/.S3261

                         1     1     1   3                              3    1     1     1   3
                     [ -(-), -(-), -(-), - ]      [   1     1     ]   [ -, -(-), -(-), -(-), - ]
                2    |   2     2     2   2 |      | -(-), -(-)    |   | 2    2     2     2   2 |
         2 Ga(1)  (Ga|                     | -  F |   2     2 ; 1 | Ga|                        |)
                     |      3   1  1  1    |   2 1|               |   |      3                 |
                     [    -(-), -, -, -    ]      [     0         ]   [    -(-), 1, 1, 1, 0    ]
                            2   2  2  2                                      2
Out[22]= ----------------------------------------------------------------------------------------
                                                      3 2
                                                 3 Ga(-)
                                                      2

In[23]:= SF/.S3261/.SListe

Be sure to apply "FOrdne" before using the following information!

                           1     1     1   3               3    1     1     1   3
                       [ -(-), -(-), -(-), - ]           [ -, -(-), -(-), -(-), - ]
                  2    |   2     2     2   2 |           | 2    2     2     2   2 |
           2 Ga(1)  (Ga|                     | - S2103 Ga|                        |)
                       |      3   1  1  1    |           |      3                 |
                       [    -(-), -, -, -    ]           [    -(-), 1, 1, 1, 0    ]
                              2   2  2  2                       2
Out[23]= {{-------------------------------------------------------------------------}}
                                                3 2
                                           3 Ga(-)
                                                2

In[33]:= R = SF/.S3261/.S2103/.Gzerl

                        1  3    3            1  3    3 2
                   Ga(-(-))  Ga(-)      Ga(-(-))  Ga(-)
                2       2       2            2       2
         2 Ga(1)  (--------------- - ----------------------)
                        3      1 3        3      1 2      2
                   Ga(-(-)) Ga(-)    Ga(-(-)) Ga(-)  Ga(1)
                        2      2          2      2
Out[33]= ---------------------------------------------------
                                   3 2
                              3 Ga(-)
                                   2

In[42]:= Simplify[R]

                1  3       2      1     3
         2 Ga(-(-))  (Ga(1)  - Ga(-) Ga(-))
                2                 2     2
Out[42]= ----------------------------------
                     3      1 3    3
              3 Ga(-(-)) Ga(-)  Ga(-)
                     2      2      2

In[52]:= ?S3261
Summation formula (Slater, Appendix (III.31)) in form of a rule.

In[53]:= ?S2103
Summation formula (Slater, Appendix (III.3)) in form of a rule.
Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ The credit for the answer should go also to Johannes Trost (mathoverflow.net/users/37436/johannes-trost) $\endgroup$
    – Zurab Silagadze
    Commented Aug 1, 2016 at 13:53
  • 1
    $\begingroup$ @Zurab Silagadze Thank you that you consider me for credit. It's okay for me. I should have posted as answer. $\endgroup$
    – Johannes Trost
    Commented Aug 1, 2016 at 14:07
10
$\begingroup$

Maple also gets $4 - 8/\pi$. This comes from an explicit formula for the partial sums:

$$ \sum_{n=1}^N a_n = {\frac { \left( 4\,N+2 \right) \left( \Gamma \left( N+1 \right) \right) ^{2}}{ \left( \Gamma \left( N+3/2 \right) \right) ^{2}}}-\frac{8} {\pi} $$

which is readily verified by mathematical induction.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Quite charming that this is summable in human terms! :) $\endgroup$
    – paul garrett
    Commented Aug 1, 2016 at 23:01
  • $\begingroup$ Yes, quite charming! $\endgroup$
    – Zurab Silagadze
    Commented Aug 2, 2016 at 4:37
4
$\begingroup$

A more general identity is furnished by $$\sum_{n=1}^N\frac{\Gamma(n)\Gamma(n+k)}{\Gamma(n+\frac12)\Gamma(n+k+\frac32)} =\frac4{2k+1}\left[\frac{\Gamma(N+1)\Gamma(N+k+1)}{\Gamma(N+\frac12)\Gamma(N+k+\frac32)}-\frac{\Gamma(k+1)}{\Gamma(k+\frac32)\sqrt{\pi}}\right],$$ which is provable by induction on $N$. Now, taking the limit $N\rightarrow\infty$ leads to $$\sum_{n=1}^{\infty}\frac{\Gamma(n)\Gamma(n+k)}{\Gamma(n+\frac12)\Gamma(n+k+\frac32)}=\frac4{2k+1}\left[1-\frac{\Gamma(k+1)}{\Gamma(k+\frac32)\sqrt{\pi}}\right].$$ To get the solution to the OP's original problem, take $k=0$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Can be generalized even further. Let $$a_n=\frac{\Gamma(n+m)\Gamma(n+k)}{\Gamma\left(n+m+1/2\right)\Gamma\left(n+k+3/2\right)}.$$ Then $$\frac{a_n}{a_{n-1}}=\frac{(n-1+m)(n-1+k)}{(n+m)(n+k)+\frac{1}{2}(m-k)-\frac{1}{4}}$$ and $$\frac{4(n+m)(n+k)}{2(k-m)+1}a_n=\frac{4(n-1+m)(n-1+k)}{2(k-m)+1}a_{n-1}+a_n,$$ which implies $$\sum_{i=1}^na_i=\frac{4(n+m)(n+k)}{2(k-m)+1}a_n-\left[\frac{4(m+1)(k+1)}{2(k-m)+1}-1\right]a_1.$$ Taking the limit, we get $$\sum\limits_{n=1}^\infty a_n=\frac{4}{2(k-m)+1}\left[1-\frac{\Gamma(m+1)\Gamma(k+1)}{\Gamma\left(m+1/2\right)\Gamma\left(k+3/2\right)}\right].$$ $\endgroup$
    – Zurab Silagadze
    Commented Apr 10, 2017 at 5:41
2
$\begingroup$

Incidentally, note that the series telescopes, as $$a_n={\Gamma(n)^2\over\Gamma(n+1/2)\Gamma(n+3/2)}=-4\big[(n-1/2)(n+1/2)-n^2\big]{\Gamma(n)^2\over\Gamma(n+1/2)\Gamma(n+3/2)}=$$ $$=-{4\,\Gamma(n)^2\over\Gamma(n-1/2)\Gamma(n+1/2)}+{4\,\Gamma(n+1)^2\over\Gamma(n+1/2)\Gamma(n+3/2)},$$ whence the partial and the limit sum given in previous answers.

Improve this answer
$\endgroup$
1
$\begingroup$

Mathematica says $$\frac{4 (\pi -2)}{\pi },$$ which is borne out by numerical approximation.

Improve this answer
$\endgroup$

You must log in to answer this question.

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