# The sum of an hydrogen atom related infinite series

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?

• Mathematica Programming Lab ( lab.wolframcloud.com ) gives $4(\pi-2)/\pi$. Aug 1, 2016 at 12:08
• 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. Aug 1, 2016 at 13:07
• 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. Aug 1, 2016 at 13:18
• Ah, sorry, by $\ll$ I meant "is less than a positive constant multiple of"... Aug 1, 2016 at 13:30

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.

• The credit for the answer should go also to Johannes Trost (mathoverflow.net/users/37436/johannes-trost) Aug 1, 2016 at 13:53
• @Zurab Silagadze Thank you that you consider me for credit. It's okay for me. I should have posted as answer. Aug 1, 2016 at 14:07

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.

• Quite charming that this is summable in human terms! :) Aug 1, 2016 at 23:01
• Yes, quite charming! Aug 2, 2016 at 4:37

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 $k=0$.

• 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 $$\sum\limits_{n=1}^\infty a_n=\frac{4}{2(k-m)+1}\left[1-\frac{\Gamma(m+1)\Gamma(k+1)}{\Gamma\left(m+1/2\right)\Gamma\left(k+3/2\right)}\right].$$ Apr 10, 2017 at 5:41

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.

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