Limit as $x\to\infty$ of $(x+3)^x/f_1(x)$, where $f_1(x)=\sum\limits_{n=3}^{x+2}n^x$

Question

So take the function $$f_{1}(x)=\sum_{n=3}^{x+2}n^x.$$ This will give results like $3^1$, for $x=1$, $3^2+4^2$, for $x=2$, and $3^3+4^3+5^3$, for $x=3$, and etc. Now, also, take the function $$f_2(x)=(x+3)^x.$$ This will give results like $4^1$, for $x=1$, $5^2$, for $x=2$, and $6^3$, for $x=3$, and etc.
Both of these kind of relate to Fermat's last theorem and Euler's related conjecture.
What's cool is that $f_1(x)=f_2(x)$, when $x=2,3$.
However, both functions separate at higher values, for example: $$f_1(4)=3^4+4^4+5^4+6^4=2258\neq f_2(4)=7^4=2401.$$ What is interesting, though, is that $f_2(x)\ge f_1(x)$, for all non-negative integer $x$ ($f_1$ wouldn't even make sense with any other domain).
Even more interesting is the function $\frac{f_2(x)}{f_1(x)}$, which appears to converge, as $x$ approaches infinity.
Her is a table of values for $x$, $f_1(x)$, $f_2(x)$, and $\frac{f_2(x)}{f_1(x)}$ (values are 5 digit approximations) $$ \begin{array}{c|lcr} x & f_1(x) & f_2(x) & \frac{f_2(x)}{f_1(x)} \\ \hline 1 & 3 & 4 & 1.3333 \\ 2 & 25 & 25 & 1 \\ 3 & 216 & 216 & 1 \\ 4 & 2258 & 2401 & 1.0633 \\ 5 & 28975 & 32768 & 1.1309 \\ 10 & 1.0277\times 10^{11} & 1.3786\times 10^{11} & 1.3414 \\ 20 & 1.1446\times 10^{27} & 1.7162\times 10^{27} & 1.4994 \\ 30 & 2.2970\times 10^{45} & 3.5927\times 10^{45} & 1.5641 \\ 40 & 1.3640\times 10^{65} & 2.1814\times 10^{65} & 1.5993 \\ 50 & 1.0090\times 10^{86} & 1.6360\times 10^{86} & 1.6214 \\ 100 & 1.1521\times 10^{201} & 1.921\times 10^{201} & 1.6681 \\ 142 & 4.8786\times 10^{306} & 8.2083\times 10^{306} & 1.6825 \\ \end{array} $$ So the question comes down to:$$\text{What is} \lim_{x\to \infty} \frac{(x+3)^x}{\sum_{n=3}^{x+2}n^x}?$$

Answer 1

Not an answer but just for the fun of playing with huge numbers and illimited precision.

Let $x=10 ^k$ and compute the expression as you did. This gives $$\left( \begin{array}{cc} k & \text{result} \\ 1 & 1.341438740 \\ 2 & 1.668056114 \\ 3 & 1.713087849 \\ 4 & 1.717760654 \\ 5 & 1.718229693 \\ 6 & 1.718276615 \\ 7 & 1.718281307 \end{array} \right)$$ which looks like $(e-1) \approx 1.718281828$.

My computer (and I) gave up for $k=8$. However, using Aitken acceleration $$p_8= p_5-\frac{(p_6-p_5)^2}{p_7-2 p_6+p_5}$$and more decimal figures for the previous calculations, the next term would be $1.71828182848$ while $(e-1) \approx 1.718281828459$.

Answer 2

0. Opening: Note that $$f_1(x)=\sum_{n=3}^{x+2}n^x=(x+2)^x\sum_{k=0}^{x-1}\left(1-\frac k{x+2}\right)^x$$ 1. Upper bound: For every $t$, $$1-t\leqslant e^{-t}$$ hence, for every $k$, $$\left(1-\frac k{x+2}\right)^x\leqslant e^{-kx/(x+2)}$$ which implies $$f_1(x)\leqslant(x+2)^x\sum_{k=0}^\infty e^{-kx/(x+2)}=\frac{(x+2)^x}{1-e^{-x/(x+2)}}$$ hence $$\frac{(x+3)^x}{f_1(x)}\geqslant\left(1+\frac1{x+2}\right)^x(1-e^{-x/(x+2)})=g(x)$$ 2. Lower bound: On the other hand, for every $t$ in $(0,\frac12)$, $$1-t\geqslant e^{-t-t^2}$$ Fix some positive $a$. Then, for every $k\leqslant a\sqrt x$ and every $x$ large enough, $$\left(1-\frac k{x+2}\right)^x\geqslant e^{-kx/(x+2)}e^{-a^2}$$ which implies $$f_1(x)\geqslant(x+2)^xe^{-a^2}\sum_{k=0}^{a\sqrt x}e^{-kx/(x+2)}\geqslant(x+2)^xe^{-a^2}\frac{1-e^{-a\sqrt xx/(x+2)}}{1-e^{-x/(x+2)}}$$ hence $$\frac{(x+3)^x}{f_1(x)}\leqslant\left(1+\frac1{x+2}\right)^xe^{a^2}\frac{1-e^{-x/(x+2)}}{1-e^{-a\sqrt xx/(x+2)}}=h_a(x)$$ 3. Cauda: Now, $$\lim_{x\to\infty} g(x)=e(1-e^{-1})=e-1$$ and $$\lim_{x\to\infty} h_a(x)=e\,e^{a^2}\frac{1-e^{-1}}{1}=(e-1)e^{a^2}$$ hence, for every positive $a$, $$e-1\leqslant\liminf_{x\to\infty}\frac{(x+3)^x}{f_1(x)}\leqslant\limsup_{x\to\infty}\frac{(x+3)^x}{f_1(x)}\leqslant(e-1)e^{a^2}$$ which, in the limit $a\to0$, implies

$$\lim_{x\to\infty}\frac{(x+3)^x}{f_1(x)}=e-1$$

Answer 3

Not an answer but just for fun and to explore others ways

By Cesaro-Stolz

$$\frac{a_x}{b_x}=\frac{(x+3)^x}{\sum_{n=3}^{x+2}n^x}\implies \frac{a_{x+1}-a_x}{b_{x+1}-b_x}=\frac{(x+4)^{x+1}-(x+3)^x}{\sum_{n=3}^{x+3}n^{x+1}-\sum_{n=3}^{x+2}n^x} =\frac{(x+4)^{x+1}-(x+3)^x}{(x+3)^{x+1}+\sum_{n=3}^{x+2}\left(n^{x+1}-n^x\right)} =\frac{(1+\frac1{x+3})^{x+1}-\frac1{x+3}}{1+\frac{\sum_{n=3}^{x+2}\left(n^{x+1}-n^x\right)}{(x+3)^{x+1}}}$$

then if $\lim_{x\to\infty} \frac{\sum_{n=3}^{x+2}\left(n^{x+1}-n^x\right)}{(x+3)^{x+1}}=L_2 \implies L_1=\lim_{x\to \infty} \frac{(x+3)^x}{\sum_{n=3}^{x+2}n^x}=\frac{e}{1+L_2}$

then again by Cesaro-Stolz we obtain

$$\frac{a_x}{b_x}=\frac{\sum_{n=3}^{x+2}\left(n^{x+1}-n^x\right)}{(x+3)^{x+1}}=\frac{\sum_{n=3}^{x+2} n^x(n-1)}{(x+3)^{x+1}}\implies \frac{a_{x+1}-a_x}{b_{x+1}-b_x} =\frac{\sum_{n=3}^{x+3} n^{x+1}(n-1)-\sum_{n=3}^{x+2} n^x(n-1)}{(x+4)^{x+2}-(x+3)^{x+1}} =\frac{(x+3)^{x+1}(x+2)+\sum_{n=3}^{x+2} n^{x}(n-1)^2}{(x+4)^{x+2}-(x+3)^{x+1}} =\frac{1+\frac{\sum_{n=3}^{x+2} n^{x}(n-1)^2}{(x+3)^{x+1}(x+2)}}{\frac{x+4}{x+2}(1+\frac1{x+3})^{x+1}-\frac1{x+2}}$$

thus if $\lim_{x\to\infty} \frac{\sum_{n=3}^{x+2} n^{x}(n-1)^2}{(x+3)^{x+1}(x+2)}=L_3 \implies L_2=\frac{1+L_3}{e}$

then again by Cesaro-Stolz we obtain

$$\frac{a_x}{b_x}=\frac{\sum_{n=3}^{x+2} n^{x}(n-1)^2}{(x+3)^{x+1}(x+2)}\implies \frac{a_{x+1}-a_x}{b_{x+1}-b_x} =\frac{\sum_{n=3}^{x+3} n^{x+1}(n-1)^2-\sum_{n=3}^{x+2} n^{x}(n-1)^2}{(x+4)^{x+2}(x+3)-(x+3)^{x+1}(x+2)} =\frac{(x+3)^{x+1}(x+2)^2+\sum_{n=3}^{x+2} n^{x}(n-1)^3}{(x+4)^{x+2}(x+3)-(x+3)^{x+1}(x+2)} =\frac{1+\frac{\sum_{n=3}^{x+2} n^{x}(n-1)^3}{(x+3)^{x+1}(x+2)^2}}{\frac{(x+4)(x+3)}{(x+2)^2}(1+\frac1{x+3})^{x+1}-\frac1{x+2}} $$

thus if $\lim_{x\to\infty} \frac{\sum_{n=3}^{x+2} n^{x}(n-1)^3}{(x+3)^{x+1}(x+2)^2}=L_4 \implies L_3=\frac{1+L_4}{e}$

If we assume for true that $L_2$, $L_3$ and $L_4$ exist and $L_2=L_3=L_4$ we obtain that

$$L_2=\frac{1+L_2}{e}\implies eL_2-L_2=1 \implies L_2=\frac{1}{e-1}$$

and then

$$L_1=\lim_{x\to \infty} \frac{(x+3)^x}{\sum_{n=3}^{x+2}n^x}=\frac{e}{1+\frac{1}{e-1}}=\frac{e(e-1)}{e-1+1}=e-1$$