Cfrgtkky

My professor recently asked to approximate this infinite sum:

$$sum_{n=1}^infty{3n[4+(1/n)]^{-n}}$$

His solution was 1.060000271. And this is what the students should come up with.

But I was asked how this could be solved to an exact value, so I asked my professor and his answer was basically "I don't know".

I simplified this formula to

$$ 3timessum_{n=1}^infty{n^{n+1}over{(4n+1)^n}}$$

But I don't see any further way to simplify that sum.

Challenge accepted. The Lagrange-Buhrmann inversion theorem gives

$$-W_0(-x)=sum_{ngeq 1}frac{n^{n-1}}{n!}x^n $$
$$sum_{ngeq 1}frac{n^n}{n!}x^n = -frac{W_0(-x)}{1+W_0(-x)} $$
$$sum_{ngeq 1}frac{n^{n+1}}{n!} x^n = -frac{W_0(-x)}{(1+W_0(-x))^3}$$
$$sum_{ngeq 1}frac{n^{n+2}}{n!} x^n = frac{W_0(-x)(2W_0(-x)-1)}{(1+W_0(-x))^5}$$

for any $x$ such that $|x|<frac{1}{e}$. If we replace $x$ with $z e^{-4z}$ we get that for any $zgeq 0$
$$sum_{ngeq 1}frac{n^{n+1}}{(n-1)!} z^{n-1} e^{-(4n+1)z} = frac{W_0(-z e^{-4z})left(2W_0(-z e^{-4z})-1right)e^{-z}}{zleft(1+W_0(-z e^{-4z})right)^5} $$
holds, and by integrating both sides over $mathbb{R}^+$ we have
$$ sum_{ngeq 1}frac{n^{n+1}}{(4n+1)^n}=int_{0}^{+infty}frac{W_0(-z e^{-4z})left(2W_0(-z e^{-4z})-1right)e^{-z}}{zleft(1+W_0(-z e^{-4z})right)^5},dz. $$
Now the integrand function in the RHS might appear as a nightmare, but its graph is actually pretty nicely gaussian-shaped, so the numerical evaluation of the RHS is fairly simple through Gaussian quadrature based on Laguerre or Hermite polynomials. Up to six figures the wanted constant is $0.353333$, remarkably close to $frac{1}{3}+frac{1}{50}=frac{53}{150}$.

A possible way for an approximation.

Consider, as you wrote,
$$sum_{n=1}^infty{n^{n+1}over{(an+epsilon)^n}}$$ ( where $a>0$) and use a Taylor expansion of the summand around $epsilon=0$. this would give
$${n^{n+1}over{(an+epsilon)^n}}=n a^{-n}-n a^{-n-1}epsilon+frac{1}{2} (n+1) a^{-n-2}epsilon ^2+Oleft(epsilon ^3right)$$
Computing the sums
$$sum_{n=1}^infty n a^{-n}=frac{a}{(a-1)^2}$$
$$sum_{n=1}^infty n a^{-n-1}=frac 1a sum_{n=1}^infty n a^{-n}=frac{1}{(a-1)^2}$$
$$sum_{n=1}^infty (n+1) a^{-n-2}=frac 1{a^2}sum_{n=1}^infty n a^{-n}+frac 1{a^2}sum_{n=1}^infty a^{-n}=frac{2 a-1}{(a-1)^2 a^2}$$ Limited to these terms, we should have
$$sum_{n=1}^infty{n^{n+1}over{(an+epsilon)^n}}approxfrac{2 a^3-2 a^2 epsilon +(2 a-1) epsilon ^2}{2 (a-1)^2 a^2}$$
Using $a=4$ and $epsilon=1$, this would give $frac{103}{288}$ and then, multiplied by $3$, $frac{103}{96}approx 1.07292$.

We could contiue with the expansion but the problem is that the next term would be
$$-frac{(n+1) (n+2) a^{-n-3}}{6 n} epsilon ^3$$ the summation of which being more difficult
$$sum_{n=1}^infty frac{(n+1) (n+2) a^{-n-3}}{6 n}=frac{4 a-3-2 (a-1)^2 log left(frac{a-1}{a}right)}{6 (a-1)^2 a^3}$$ making for your case
$$3sum_{n=1}^infty{n^{n+1}over{(an+epsilon)^n}}approx frac{103}{96}-frac{13+18 log left(frac{4}{3}right)}{1152}=frac{1223-18 log left(frac{4}{3}right)}{1152}approx 1.05714$$

We could continue that way but the next term would be
$$frac{(n+1) (n+2) (n+3) a^{-n-4}}{24 n^2} epsilon ^4$$ the summation of which making appearing the polylogarithm function (don't worry : sooner or later, you will learn about it !).

So, I shall stop here and conlude that the value of your summation is somewhere between the two numerical values given above. Notice that taking the mean of both, you have $approx 1.06505$

Edit

For the specific case $(a=4,epsilon=1)$, I put below the expression of the infinite sum and its numerical representation
$$left(
begin{array}{ccc}
p & text{formula} & text{value} \
1 & 1 & 1.00000 \
2 & frac{103}{96} & 1.07292 \
3 & frac{1223-18 log left(frac{4}{3}right)}{1152} & 1.05714 \
4 & frac{6530-63 log left(frac{4}{3}right)+18
text{Li}_2left(frac{1}{4}right)}{6144} & 1.06066 \
5 & frac{391766-4095 log left(frac{4}{3}right)+630
text{Li}_2left(frac{1}{4}right)-216
text{Li}_3left(frac{1}{4}right)}{368640} & 1.05984 \
6 & frac{9402433-97515 log left(frac{4}{3}right)+17145
text{Li}_2left(frac{1}{4}right)-2718
text{Li}_3left(frac{1}{4}right)+1080
text{Li}_4left(frac{1}{4}right)}{8847360} & 1.06004 \
7 & frac{87756019-910665 log left(frac{4}{3}right)+157815
text{Li}_2left(frac{1}{4}right)-30240
text{Li}_3left(frac{1}{4}right)+4788
text{Li}_4left(frac{1}{4}right)-2160
text{Li}_5left(frac{1}{4}right)}{82575360} & 1.05999
end{array}
right)$$