Symbolic solution to the infinite sum of $\sum_{n=1}^\infty{3n[4+(1/n)]^{-n}}$

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

$$3\times\sum_{n=1}^\infty{n^{n+1}\over{(4n+1)^n}}$$

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

• "But I was asked how this could be solved to an exact value" - You're not being asked to find the infinite sum, you're being asked to approximate it. You could, hypothetically, replace that $\infty$ sign with $10,000$ or whatever, i.e. taking the first $10,000$ terms. – Eevee Trainer Jan 2 '19 at 4:21
• Ohh, disclosure, I'm one of the tutors in this class. So other students ask me, as they should. And if I can't solve it, I will ask the professor. And if that doesn't work and I ponder a while on the problem, well, now I'm here. – Johannes Kuhn Jan 2 '19 at 4:23
• Oh, I see. And looking at the problem I misread anyhow. My bad. Of course it's also worth keeping in mind that not all sums are able to be evaluated easily: a summation with a similar thing (the index raised to itself): en.wikipedia.org/wiki/Sophomore%27s_dream – Eevee Trainer Jan 2 '19 at 4:25
• (Of course that's not to say this doesn't converge or that we can't show the sum converges to some finite value. Just worth keeping in mind that it might not be possible, or, at best, very nontrivial.) – Eevee Trainer Jan 2 '19 at 4:26
• The sum of this series is unlikely to have a closed form expression. – Robert Israel Jan 2 '19 at 4:28

Challenge accepted. The Lagrange-Buhrmann inversion theorem gives

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

for any $$x$$ such that $$|x|<\frac{1}{e}$$. If we replace $$x$$ with $$z e^{-4z}$$ we get that for any $$z\geq 0$$ $$\sum_{n\geq 1}\frac{n^{n+1}}{(n-1)!} z^{n-1} e^{-(4n+1)z} = \frac{W_0(-z e^{-4z})\left(2W_0(-z e^{-4z})-1\right)e^{-z}}{z\left(1+W_0(-z e^{-4z})\right)^5}$$ holds, and by integrating both sides over $$\mathbb{R}^+$$ we have $$\sum_{n\geq 1}\frac{n^{n+1}}{(4n+1)^n}=\int_{0}^{+\infty}\frac{W_0(-z e^{-4z})\left(2W_0(-z e^{-4z})-1\right)e^{-z}}{z\left(1+W_0(-z e^{-4z})\right)^5}\,dz.$$ 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 $${n^{n+1}\over{(an+\epsilon)^n}}=n a^{-n}-n a^{-n-1}\epsilon+\frac{1}{2} (n+1) a^{-n-2}\epsilon ^2+O\left(\epsilon ^3\right)$$ 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 $$\sum_{n=1}^\infty{n^{n+1}\over{(an+\epsilon)^n}}\approx\frac{2 a^3-2 a^2 \epsilon +(2 a-1) \epsilon ^2}{2 (a-1)^2 a^2}$$ 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 $$3\sum_{n=1}^\infty{n^{n+1}\over{(an+\epsilon)^n}}\approx \frac{103}{96}-\frac{13+18 \log \left(\frac{4}{3}\right)}{1152}=\frac{1223-18 \log \left(\frac{4}{3}\right)}{1152}\approx 1.05714$$

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 $$\left( \begin{array}{ccc} p & \text{formula} & \text{value} \\ 1 & 1 & 1.00000 \\ 2 & \frac{103}{96} & 1.07292 \\ 3 & \frac{1223-18 \log \left(\frac{4}{3}\right)}{1152} & 1.05714 \\ 4 & \frac{6530-63 \log \left(\frac{4}{3}\right)+18 \text{Li}_2\left(\frac{1}{4}\right)}{6144} & 1.06066 \\ 5 & \frac{391766-4095 \log \left(\frac{4}{3}\right)+630 \text{Li}_2\left(\frac{1}{4}\right)-216 \text{Li}_3\left(\frac{1}{4}\right)}{368640} & 1.05984 \\ 6 & \frac{9402433-97515 \log \left(\frac{4}{3}\right)+17145 \text{Li}_2\left(\frac{1}{4}\right)-2718 \text{Li}_3\left(\frac{1}{4}\right)+1080 \text{Li}_4\left(\frac{1}{4}\right)}{8847360} & 1.06004 \\ 7 & \frac{87756019-910665 \log \left(\frac{4}{3}\right)+157815 \text{Li}_2\left(\frac{1}{4}\right)-30240 \text{Li}_3\left(\frac{1}{4}\right)+4788 \text{Li}_4\left(\frac{1}{4}\right)-2160 \text{Li}_5\left(\frac{1}{4}\right)}{82575360} & 1.05999 \end{array} \right)$$