Find the summation of $\sum_{n\geq1}\frac{3^n}{n\left(\frac{1}{n}+1\right)^n}x^n$

Question

I was trying to find what the summation of $$\sum_{n\geq1}\frac{3^n}{n\left(\frac{1}{n}+1\right)^n}x^n$$ is, but I'm kind of stuck.

I recognized the pattern at the bottom as $$\lim_{n\to+\infty}\left(1+\frac{1}{n}\right)^n = e$$ but I don't know how this should help. The only thing I would do is to group those $3$ and $x$ and put them under the same exponent. Then, if I had $$\frac{(3x)^n}{n}$$ only, I would be able to find the result as $-\log(1-3x)$ but there is also the other part I stated before so I cannot conclude anything.

What am I missing? Thank you all in advance

Answer 1

Let first $x=\frac t 3$ and simplify to write $$S=\sum_{n=1}^\infty\frac {n^{n-1} } {(n+1)^n }\,t^n$$

For large values of $n$ $$\frac {n^{n-1} } {(n+1)^n }=\frac{1}{e n}+\frac{1}{2 e n^2}-\frac{5}{24 e n^3}+\frac{5}{48 e n^4}-\frac{337}{5760 e n^5}+O\left(\frac{1}{n^6}\right)$$ The terms alternate after the second one.

So, you can have as simplest bounds in terms of polylogarithms $$-\frac{\log (1-t)}{e}+\frac{\text{Li}_2(t)}{2 e}-\frac{5 \text{Li}_3(t)}{24 e}<S<-\frac{\log (1-t)}{e}+\frac{\text{Li}_2(t)}{2 e}$$

$$\left( \begin{array}{cccc} t & \text{left bound} & S & \text{right bound} \\ 0.1 & 0.049873 & 0.052374 & 0.057636 \\ 0.2 & 0.105166 & 0.110209 & 0.120902 \\ 0.3 & 0.167259 & 0.174886 & 0.191202 \\ 0.4 & 0.238153 & 0.248412 & 0.270563 \\ 0.5 & 0.320919 & 0.333860 & 0.362092 \\ 0.6 & 0.420639 & 0.436318 & 0.470917 \\ 0.7 & 0.546723 & 0.565202 & 0.606509 \\ 0.8 & 0.719986 & 0.741336 & 0.789777 \\ 0.9 & 1.005700 & 1.030000 & 1.086140 \\ \end{array} \right)$$

Doing the same with the next two terms $$\left( \begin{array}{cccc} t & \text{left bound} & S & \text{right bound} \\ 0.1 & 0.051570 & 0.052374 & 0.053729 \\ 0.2 & 0.108597 & 0.110209 & 0.112930 \\ 0.3 & 0.172464 & 0.174886 & 0.178985 \\ 0.4 & 0.245176 & 0.248412 & 0.253900 \\ 0.5 & 0.329807 & 0.333860 & 0.340749 \\ 0.6 & 0.431444 & 0.436318 & 0.444623 \\ 0.7 & 0.559503 & 0.565202 & 0.574936 \\ 0.8 & 0.734807 & 0.741336 & 0.752514 \\ 0.9 & 1.022640 & 1.030000 & 1.042640 \\ \end{array} \right)$$

Edit

As show above, we have $$\frac {n^{n-1} } {(n+1)^n }=\frac{1}{e n}+\frac{1}{2e}\sum_{k=2}^\infty (-1)^k\, \frac{a_k}{n^k}$$ where the first $a_k$ form the sequence $$\left\{1,\frac{5}{12},\frac{5}{24},\frac{337}{2880},\frac{137}{1920},\frac{67177}{1451520},\frac{18289}{580608},\frac{15527801}{696729600},\cdots\right\}$$ This gives $$S=-\frac 1e \log(1-t)+\frac{1}{2e}\sum_{k=2}^\infty (-1)^k\, {a_k}\,\text{Li}_k(t)$$

As the simplest approximation, using Padé approximant $$\text{Li}_k(t) \sim \frac{2^k\, t}{2^k-t}$$

To show its validity, computing the norm $$\Phi_k=\int_0^1 \left(\text{Li}_k(t)-\frac{2^k t}{2^k-t}\right)^2\,dt$$ A few numbers $$\left( \begin{array}{cc} k & \Phi_k \\ 2 & 5.87884\times 10^{-3} \\ 3 & 3.41844\times 10^{-4} \\ 4 & 2.84543\times 10^{-5} \\ 5 & 2.75109\times 10^{-6} \\ 6 & 2.85300\times 10^{-7} \\ 7 & 3.06285\times 10^{-8} \\ 8 & 3.34627\times 10^{-9} \\ 9 & 3.68878\times 10^{-10} \\ 10 & 4.08488\times 10^{-11} \\ \end{array} \right)$$

So, as a first approximation, $$S \sim -\frac 1e \log(1-t)+\frac{1}{2e}\sum_{k=2}^\infty (-1)^k\, {a_k}\,\frac{2^k t}{2^k-t}$$

Using only the $a_k$ given in the table $$\left( \begin{array}{ccc} t & \text{approximation} & \text{value of } S \\ 0.1 & 0.052194 & 0.052374 \\ 0.2 & 0.109790 & 0.110209 \\ 0.3 & 0.174085 & 0.174886 \\ 0.4 & 0.246953 & 0.248412 \\ 0.5 & 0.331265 & 0.333860 \\ 0.6 & 0.431778 & 0.436318 \\ 0.7 & 0.557333 & 0.565202 \\ 0.8 & 0.727648 & 0.741336 \\ 0.9 & 1.005370 & 1.030000 \\ \end{array} \right)$$