Csdrhrt

By the comparison test, the series converges.
$$sum_{n=0}^{infty}frac{(2n+15)^2}{(2n+11)^3sqrt{(2n+13)}}=l$$
$$lsimeq0.39483198670640570922670209458869656945532664947383304288146572043505953641$$
(74 digits displayed)

I would like to know if it is possible to find a closed-form expression for $l$.

Edit

I used a PARI/GP function (sumnum) to compute an approximate value of $l$:

 gp > #

    timer = 1 (on)

 gp > p 74

    realprecision = 77 significant digits (74 digits displayed)

 gp > sumnum(n=0, (2*n+15)^2/(2*n+11)^3/sqrt(2*n+13),sumnuminit([+oo,-1.5]))

 time = 15 ms.

 %1 = 0.39483198670640570922670209458869656945532664947383304288146572043505953641

 gp > sum(k=0,97,binomial(-1/2,k)*(zetahurwitz(k+3/2,11/2)+4*zetahurwitz(k+5/2,11/2)+4*zetahurwitz(k+7/2,11/2)))/2/sqrt(2)

 time = 62 ms.

 %2 = 0.39483198670640570922670209458869656945532664947383304288146572043505953641

 gp > p 400

    realprecision = 404 significant digits (400 digits displayed)

 gp > sumnum(n=0, (2*n+15)^2/(2*n+11)^3/sqrt(2*n+13),sumnuminit([+oo,-1.5]))

 time = 1,078 ms.

 %3 = 0.3948319867064057092267020945886965694553266494738330428814657204350595364122729251535099721677148100865638904442238183504239633022203492259575689525638233484603909542517059855919506733374353926795281971622814159109637670427399334579124116308394274900167149611375839262362929064148759669151317574581119034347857659144521214162698550204609725028401793070583608454779328237780415473232520384318559950198

 gp > sum(k=0,537,binomial(-1/2,k)*(zetahurwitz(k+3/2,11/2)+4*zetahurwitz(k+5/2,11/2)+4*zetahurwitz(k+7/2,11/2)))/2/sqrt(2)

 time = 3,532 ms.

 %4 = 0.3948319867064057092267020945886965694553266494738330428814657204350595364122729251535099721677148100865638904442238183504239633022203492259575689525638233484603909542517059855919506733374353926795281971622814159109637670427399334579124116308394274900167149611375839262362929064148759669151317574581119034347857659144521214162698550204609725028401793070583608454779328237780415473232520384318559950198

Not a closed form but a rather good approximation of it.

Rewriting as $$a_n=frac{(2n+15)^2}{(2n+11)^3sqrt{2n+13}}=frac{(2n+15)^2}{(2n+11)^3sqrt{2n+11+2}}=frac{(2n+15)^2}{(2n+11)^{frac 72}sqrt{1+frac 2{2n+11}}}$$ we can write
$$frac 1{sqrt{1+frac 2{2n+11}}}=sum_{k=0}^infty binom{-frac{1}{2}}{k}frac{2^k}{(2n+11)^k}$$ and then face summations of terms
$$S_k=sum_{n=0}^inftyfrac{(2n+15)^2}{(2n+11)^{k+frac 72}}=2^{-k-frac{3}{2}} left(zeta left(k+frac{3}{2},frac{11}{2}right)+4 zeta
left(k+frac{5}{2},frac{11}{2}right)+4 zeta
left(k+frac{7}{2},frac{11}{2}right)right)$$ and then for the considered summation
$$Sigma=sum_{n=0}^infty frac{(2n+15)^2}{(2n+11)^3sqrt{2n+13}}$$ So, the partial sums
$$Sigma_p=frac 1{2sqrt 2}sum_{k=0}^p binom{-frac{1}{2}}{k} left(zeta left(k+frac{3}{2},frac{11}{2}right)+4 zeta
left(k+frac{5}{2},frac{11}{2}right)+4 zeta
left(k+frac{7}{2},frac{11}{2}right)right)$$ Computing
$$left(
begin{array}{cc}
p & Sigma_p \
5 & 0.39483148307398830445048206504782536496698034745960346801382376024 \
10 & 0.39483198676616324735777911451436673248113391244073226133701558847 \
15 & 0.39483198670639658527502248284985700122310258168712958250026452734 \
20 & 0.39483198670640571076206255275333399438974388443990146144695977886 \
25 & 0.39483198670640570922643131344152894240169346679460070946980052257 \
30 & 0.39483198670640570922670214360763624644791582367286943018650562365 \
35 & 0.39483198670640570922670209457967764179263574295643280410747243813 \
40 & 0.39483198670640570922670209458869824722904384200792304484978282169 \
45 & 0.39483198670640570922670209458869656914069678422236918330533995116 \
50 & 0.39483198670640570922670209458869656945538601634396101492730552390 \
55 & 0.39483198670640570922670209458869656945532663821679612114844784572 \
60 & 0.39483198670640570922670209458869656945532664947597620995355367822 \
65 & 0.39483198670640570922670209458869656945532664947383263347402134364 \
70 & 0.39483198670640570922670209458869656945532664947383304295989988132 \
75 & 0.39483198670640570922670209458869656945532664947383304288145065669 \
80 & 0.39483198670640570922670209458869656945532664947383304288146572333 \
85 & 0.39483198670640570922670209458869656945532664947383304288146572043 \
90 & 0.39483198670640570922670209458869656945532664947383304288146572044 \
95 & 0.39483198670640570922670209458869656945532664947383304288146572044 \
100 & 0.39483198670640570922670209458869656945532664947383304288146572044
end{array}
right)$$

Edit

In order to know how many terms have to be added for $p$ significant digits, since we face an alternatin series, consider
$$a_k=frac 1{2sqrt 2}binom{-frac{1}{2}}{k} left(zeta left(k+frac{3}{2},frac{11}{2}right)+4 zeta
left(k+frac{5}{2},frac{11}{2}right)+4 zeta
left(k+frac{7}{2},frac{11}{2}right)right)$$ A quick and dirty linear regression (built for $10 leq k leq 1000$ by steps of $10$) shows that (and this is an almost perfect fit)
$$log_{10}(|a_k|)=-0.740989, k-2.51445$$ So, for $p$ exact digits, we need to sum up $lceil 1.35, p -3.40 rceil$ terms (just as reflected by the values in the above table). For $74$ digits as given in the post, then $k=97$ which has been verified. Using the original summation, I suppose that billions of terms have been added (summing from $n=0$ to $n=10^9$ leading to $0.39481$). This look normal since, for large values of $n$, $frac{a_{n+1}}{a_n}=1-frac{3}{2 n}+Oleft(frac{1}{n^2}right)$ which is extremely slow.