A few months ago I derived a beautiful fact:

$$ \sum_{n=k+1}^\infty n^{k-n}=\int_{0}^{1} t^{k-t}dt~~~(*) $$

for every natural $k$. Generally:

$$ \sum_{n=1}^\infty \frac{a^n}{(n+s)^n}=\int_{-s}^{a-s} \frac{a^t}{(t+s)^t}dt $$

It is easy to prove. You just need to use the fact that $$ \frac{1}{(n+c)^n}=\int_{0}^\infty \frac{t^{n-1}}{(n-1)!} e^{-t(n+c)}dt $$ I know about Sophomore's dream, but even after a long search I didn't find fact $(*)$ in the literature. Please help me and answer, is it original or not?

(+ 1 edition) After some time I derived another generalization (the fact above is just the case with $b=0$):

$$ \sum_{n=1}^\infty \frac{a^{n}}{(n+c)^{n}} \frac{\Gamma(n+b)}{\Gamma(n)(n+c)^b}=\int\limits_{-c}^{a-c} \left ( \mathrm{ln}\frac{a}{t+c}\right )^b\frac{a^t}{(t+c)^t} dt $$

(where $\Gamma(x)$ denotes the Gamma-function). Other sum is interesting too:

$$ \sum_{n=0}^\infty \frac{a^{n}\Gamma(n+b)\Gamma(n+d)}{n!(n+c)^{n+b}\Gamma(d)}=\int_{0}^\infty \frac{x^{b-1}e^{-x(c-d)}}{(e^x-ax)^{d}} dx $$

  • 1
    $\begingroup$ It's not original. Look for Euler's Gamma function and you'll find that. $\endgroup$ – FrodCube Jul 3 '15 at 17:46
  • $\begingroup$ I think that you can prove that last line from $n!=\int_0^\infty t^ne^{-t}dt$ for $n>-1$ (this is basically the definition of the Gamma function) and doing some substitution. (To prove the equation I just gave, use integration by parts and induction.) $\endgroup$ – Akiva Weinberger Jul 3 '15 at 18:22
  • $\begingroup$ I know, how to prove it, I proved it myself. I just don't know, am I the first person, who did it or not. $\endgroup$ – Danil Krotkov Jul 3 '15 at 18:30
  • $\begingroup$ @DanilKrotkov I have not seen this shifted/scaled version, but inasmuch as it is simply a shifted/scaled adaptation, I would speculate that your development is not, unfortunately, an original. Is that all you need for now or was there anything about the development that you wish to discuss further? $\endgroup$ – Mark Viola Jul 3 '15 at 18:49
  • $\begingroup$ @Dr.MV Do you know, where to find an article or a book with this fact in it? Did you see it online or not? Send me a link, please. (And sorry for my english, I know it is not good) $\endgroup$ – Danil Krotkov Jul 3 '15 at 18:55

Although the OP has stated that he/she has already derived the more general form of the so-called "Sophomore's Dream," I thought that it might benefit others to see the development herein. So, here we go ...

$$\begin{align} \int_{-s}^{a-s} \frac{a^t}{(t+s)^t}dt&=a\int_0^1 t^st^{-at}dt \tag1\\\\ &=\sum_{n=0}^{\infty}\frac{(-1)^na^{n+1}}{n!}\int_0^1t^{n+s}\log^nt\,dt \tag2\\\\ &=\sum_{n=0}^{\infty}\frac{(-1)^na^{n+1}}{n!}\frac{(-1)^n}{(n+1)^{n+1}}\int_0^{\infty} t^ne^{\frac{n+1+s}{n+1}x}dx \tag3\\\\ &=\sum_{n=0}^{\infty}\frac{a^{n+1}}{n!(n+1+s)^{n+1}}\int_0^{\infty}t^ne^{-t}dt \tag4\\\\ &=\sum_{n=0}^{\infty}\frac{a^{n+1}}{(n+1+s)^{n+1}}\tag5\\\\ &=\sum_{n=1}^{\infty}\frac{a^{n}}{(n+s)^{n}}\tag6 \end{align}$$

as was to be shown!



We enforced the substitution $t \to at-s$


We wrote $t^{-at}=e^{-at\log t}$ and used the power series representation for $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. We also used the uniform convergence of the power series to justify interchanging the integral and summation.


We enforced the substitution $t\to e^{-t/(n+1)}$.


We enforced the substitution $t\to \frac{n+1}{n+1+s}t$.


We used the integral representation of the Gamma Function $\Gamma (z)=\int_0^{\infty}t^{z-1}e^{-t}dt$, which for $z=n+1$ is $\Gamma (n+1)=n!$.


We shifted the index of summation using $n\to n-1$


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