# Possible to get a closed form expression, or an upper bound, for $f(n)=\sum_{m=1}^\infty \bigg(\frac{m+n}{3}\bigg)^{m+n}\bigg(\frac{1}{m}\bigg)^m$?

Is it possible to get a closed form expression, or an upper bound, for the following function $$f$$ which is given by an infinite summation:

$$f(n) = \sum_{m=1}^\infty \bigg(\frac{m+n}{3}\bigg)^{m+n} \bigg(\frac{1}{m}\bigg)^m,$$ for $$n > 0$$? Also, $$n$$ can be assumed to be large if this is any help.

Note when $$n=0$$, it is simply $$f(0) = \frac{1}{2}.$$

• It looks like a some sort of factorial when graphed and the sum looks similar to a binomial series. – Jam Oct 29 '19 at 11:59
• Is there anything that suggests the function has a closed form? – Jam Oct 29 '19 at 12:06
• I see. Personally I don't actually think it's that simple and there are simpler series (e.g. $\sum_{i=1}^{\infty}\frac{x^{x}}{i^{i}}$) that don't have closed forms. You can possibly get a nice asymptotic for your series though. – Jam Oct 29 '19 at 12:15
• The function growth is at least exponential. $$f(n) \geq \sum_{m=1}^{\infty} \left(\frac{m}{3}\right)^{m+n}\left(\frac{1}{m}\right)^m = \sum_{m=1}^{\infty} \frac{1}{3^m}\left(\frac{m}{3}\right)^n > \frac{1}{3^4}\left(\frac{4}{3}\right)^n.$$ – JoshuaZ Oct 29 '19 at 13:13
• Using Stirling formula we have a problem with small values of $m$ . But how is the approximation $~\displaystyle f(n)\approx n!\left(\frac{3}{2}\left(\frac{e}{2}\right)^n - \left(\frac{e}{3}\right)^n\right)~$ for you ? Especially for small or large $n$ ? – user90369 Oct 30 '19 at 15:06

We can derive bounds and approximations for $$f(n)$$ that suggest properties that such a closed form must have.

$$f(n)=\sum_{i=1}^\infty\frac{\left(\frac{n+i}{3}\right)^{n+i}}{i^i}$$

When $$n$$ is fixed, the summands, $$a_i$$, are increasing over $$(0,\frac{n}{2})$$ but decreasing over $$(\frac{n}{2},\infty)$$. This can be shown by considering their derivatives. By bounding the summand with continuous curves and accounting for the peak at $$n/2$$, we can deduce that

$$f(n)\leq\int_{1/2}^{n/2-1/2}a_{i+1/2}\,\mathrm{d}i+ a_{n/2}+\int_{n/2+1/2}^{\infty}a_{i-1/2}\,\mathrm{d}i$$

Then, by substitution with $$i\mapsto i\pm1/2$$, $$f(n)\leq\int_{1}^{\infty}a_{i}\,\mathrm{d}i+ a_{n/2}$$

Rearranging $$a_i$$ to $$\left(\frac{n+i}{3}\right)^{n}\cdot e^{-i\ln\left(\frac{3}{\frac{n}{i}+1}\right)}$$ allows us to bound it by incorporating the inequality $$1-\frac1x\leq \ln x \leq x-1$$. Hence,

$$\left(\frac{n+i}{3}\right)^{n}\cdot e^{\frac{i\left(n-2i\right)}{n+i}}\leq a_i\leq \left(\frac{n+i}{3}\right)^{n}\cdot e^{\frac{n}{3}-\frac{2}{3}i}$$

Combining this upper bound with the previous integral inequality,

\begin{aligned}f(n)&\leq\frac{e^{\frac{n}{3}}}{3^{n}}\int_{1 }^{\infty }\frac{\left(n+i\right)^{n}}{e^{\frac{2}{3}i}}\,\mathrm{d}i+ \left(\frac{n}2\right)^{n} \\ &\leq \frac{3e^{n}}{2^{n+1}}\Gamma\left(n+1,\frac{2\left(n+1\right)}{3}\right)+ \left(\frac{n}2\right)^{n}\end{aligned}

where $$\Gamma(s,x)=\int_1^\infty t^{s-1}e^{-t}\,\mathrm{d}t$$ is the incomplete gamma function. When $$x\gg 0$$, we have $$\Gamma(s,x)\approx x^{s-1}e^{-x}$$. Hence, for large $$n$$, $$f(n)\approx \frac{3}{2e^{\frac{2}{3}}}\left(\frac{e^{\frac{1}{3}}}{3}\left(n+1\right)\right)^{n}+\left(\frac{n}{2}\right)^{n}$$

### Update 2019-11-05. Improved bounds.

Expanding on @user90369's ingenious use of Stirling's approximation to bound $$n^n$$, we can use the existing upper and lower bounds for $$n!$$ to sandwich $$n^n$$. We can then choose the relevant bounds for the numerator and denominator of the summands of $$f(n)$$ to find the improved bounds

$$\frac{\sqrt{2\pi}e^{n}}{3^{n}e}\leq \frac{f(n)}{\sum_{i=1}^{\infty}\frac{\left(n+i\right)!\sqrt{\frac{i}{n+i}}}{3^{n}\cdot i!}} \leq \frac{e^{n+1}}{\sqrt{2\pi}3^{n}} \\ \\ \frac{\sqrt{2\pi}}{e\sqrt{n+1}} \leq \frac{f(n)}{\frac{e^{n}}{3^{n}}\sum_{i=1}^{\infty}\frac{\left(n+i\right)!}{3^{i}\cdot i!}} \leq \frac{e}{\sqrt{2\pi}} \\ \frac{\sqrt{2\pi}}{e\sqrt{n+1}} \leq \frac{f(n)}{e^{n}n!\left(\frac{3}{2}\cdot2^{-n}-3^{-n}\right)} \leq \frac{e}{\sqrt{2\pi}}$$

So - unless I'm mistaken, since I'm not great with asymptotic analysis - this means that $$f(n)\in\mathcal{O}\left(\sqrt{n}\left(\frac{n}2\right)^n\right)$$ and $$f(n)\in\mathcal{\Omega}\left(\left(\frac{n}2\right)^n\right)$$. This lends itself to the natural approximation $$f(n)\approx k\sqrt{n}\left(\frac{n}{2}\right)^n$$ or $$f(n)\approx k\left(\frac{n}{2}\right)^n$$, the best of which seems to be the good approximation

$$f(n)\approx 2.172\sqrt{n}\left(\frac{n}{2}\right)^n$$

• By using Stirling's approximation twice and using the approximation $\frac{\sqrt{i}}{\sqrt{n+i}}\approx \frac{\sqrt{i}}{\sqrt{n}}$ when $n$ is large and then assuming $\sqrt{i}\ll (n+i)!, i!, 3^i$, we have $f(n)\approx \frac{e^{n}n!}{\sqrt{n}}\left(\frac{3}{2}2^{-n}-3^{-n}\right)$. – Jam Nov 4 '19 at 13:23
• Very nice approximation and its great to see the techniques used in case I come across similar situations in future..thanks! – sonicboom Nov 4 '19 at 14:38
• Footnote: I think the $2.172\ldots$ constant is $\frac{\sqrt{6\pi}}{2}$, which would fit given the relation to Stirling's approximation. – Jam Nov 5 '19 at 22:50
• Footnote 2: I believe that this can be generalised to $\frac{1}{a^{n}}\sum_{i=1}^{\infty}\frac{\left(n+i\right)^{n+i}}{\left(ai\right)^{i}}\sim \frac{\sqrt{2a\pi n}}{a-1}\left(\frac{n}{a-1}\right)^{n}$. – Jam Nov 5 '19 at 23:05

Only a comment.

I am not sure if you will like this, because my commentary shows that it's unlikely that you will get a satisfying answer in relation to a closed form .

$$\sum\limits_{m=1}^\infty\left(\frac{m+n}{3}\right)^{m+n}\left(\frac{1}{m}\right)^m = \frac{1}{3^n}\sum\limits_{v=0}^\infty{\binom n v}n^{n-v}\left(\frac{d^{v+1}}{dx^{v+1}}\int\limits_0^\infty\frac{h(e^{\frac{t}{3}e^{x-t}})^n}{1-\ln h(e^{\frac{t}{3}e^{x-t}})}dt\right)\big{|}_{x=0}$$

$$h(x)~$$ is the infinite power tower . $$\,$$It's useful if $$~n\in\mathbb{N}_0$$ .

Conclusion: Lower bound and an upper bound make sense, especially an asymptotic formula.

I think the upper bound $$~\displaystyle n!\left(\frac{3}{2}\left(\frac{e}{2}\right)^n-\left(\frac{e}{3}\right)^n\right)~$$ for $$~f(n)~$$ could be a good start for looking for a better one.

• Please explain how this suggests that a closed form is unlikely. – Jam Oct 29 '19 at 14:12
• @Jam : Because of the integral, of course. And there is no need for downvotes, it's only a comment, a base for considerations, not a claim, not a proof. – user90369 Oct 29 '19 at 14:14
• Interesting answer thanks! I have derived an upper bound now..it gives the rate of convergence which is sufficient for what I need. But I would still be interested to see if a closed form is possible or an ‘almost closed form’ if not. – sonicboom Oct 29 '19 at 15:23
• @user90369 That upper bound is very nice, I would be interested to see how you derived it. – eurocoder Nov 4 '19 at 14:35
• @eurocoder : You've accepted the other answer, so I think you've tested that it is a better approximation. That's good. But to answer your question: I only used the Stirling formula $~n!\approx (n/e)^n\sqrt{2\pi n}~$ , means: $~x^x\approx x!e^x/\sqrt{2\pi x}~$ with $~x\in\{n,m+n\}~$ . – user90369 Nov 4 '19 at 16:23