I am interested in finding bounds on cumulants in terms of moments.
For example, [this paper](https://link.springer.com/content/pdf/10.1007%2FBF01043479.pdf) alludes to the bound
\begin{align}
|\kappa\_n|\le n^n E[|X-E[X]|^n]
\end{align}
where $\kappa\_n$ is the $n$-th cumulant.
However, due to the language barrier, tracking down the proof of this is difficult.


I would like to see the proof of this bound or something similar to this.



Bounds on cumulants in terms of moments
Let $k\_n:=\kappa\_n$, $a\_n:=E(X-EX)^n$, $b\_n:=E|X-EX|^n$, so that $|a\_n|\le b\_n$. We have to show that
$$|k\_n|\le n^n b\_n$$
for natural $n$. For $n=1,2$ this is obvious. The key is the recursion
$$k\_n=a\_n-\sum\_{m=1}^{n-1}\binom{n-1}{m-1}k\_m a\_{n-m}$$
at the end of [this section](https://en.wikipedia.org/wiki/Cumulant#Cumulants_and_moments) , which implies
$$|k\_n|\le b\_n+\sum\_{m=1}^{n-1}\binom{n-1}{m-1}|k\_m| b\_{n-m}.$$
This allows us to use the induction on $n$, which implies
$$|k\_n|\le l\_n b\_n,$$
where
$$l\_n:=1+\sum\_{m=1}^{n-1}\binom{n-1}{m-1}m^m;$$
here we used the log-convexity of $b\_j$ in $j$ together with the fact $b\_0=1$ (or, equivalently, Hölder's inequality), which implies $b\_m b\_{n-m}\le b\_n$.


It remains to show that $l\_n\le n^n$ for $n\ge2$. This is trivial for $n=2$. For $n\ge3$, proceed again by induction:
\begin{align\*}
l\_n&=1+\sum\_{m=1}^{n-1}\binom{n-2}{m-1}m^m+\sum\_{m=1}^{n-1}\binom{n-2}{m-2}m^m \\ 
&=1+\sum\_{m=1}^{n-2}\binom{n-2}{m-1}m^m+(n-1)^{n-1}+\sum\_{m=2}^{n-1}\binom{n-2}{m-2}m^m \\ 
&=l\_{n-1}+(n-1)^{n-1}+\sum\_{j=1}^{n-2}\binom{n-2}{j-1}(j+1)^{j+1} \\ 
&\le l\_{n-1}+(n-1)^{n-1}+\sum\_{j=1}^{n-2}\binom{n-2}{j-1}j^j c\_n \\ 
&=l\_{n-1}+(n-1)^{n-1}+(l\_{n-1}-1) c\_n \\ 
&\le (n-1)^{n-1}+(n-1)^{n-1}+(n-1)^{n-1}c\_n \\ 
&=(n-1)^{n-1}\Big(2+\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\Big)=:r\_n 
\end{align\*}
where
$$c\_n:=\max\_{1\le j\le n-2}\frac{(j+1)^{j+1}}{j^j}=\frac{(n-1)^{n-1}}{(n-2)^{n-2}},$$
because
$$\frac{(j+1)^{j+1}}{j^j}=\Big(1+\frac1j\Big)^j(j+1)$$
is increasing in $j\ge1$.


It remains to check that $r\_n\le n^n$ for all real $n>2$, which, by substitution $n=1+\frac1t$, can be rewritten as
\begin{equation}
 f\_1(t)f\_2(t)\le1, \tag{1}
\end{equation}
where
\begin{equation\*}
 f\_1(t):=e \left(\frac{1}{t+1}\right)^{\frac{1}{t}+1},\quad
 f\_2(t):=\frac{\left(\left(\frac{1}{t}\right)^{\frac{1}{t}}
 \left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}+2\right) t}{e}; 
\end{equation\*}
everywhere here, $t\in(0,1]$.


To prove (1), it is enough to prove
\begin{equation}
 f\_1\le g\_1 \tag{2},
\end{equation}
\begin{equation}
 f\_2\le g\_2 \tag{3},
\end{equation}
\begin{equation}
 g\_1g\_2\le1 \tag{4}
\end{equation}
on $(0,1]$, where
\begin{equation\*}
 g\_1(t):=\frac{7 t^2}{24}-\frac{t}{2}+1,\quad
 g\_2(t):=\left(\frac{2}{e}-\frac{1}{2}\right) t+1. 
\end{equation\*}


To prove (2), consider
\begin{equation\*}
 d\_1:=\ln f\_1-\ln g\_1 
\end{equation\*}
and then $d\_{11}(t):=t^2d\_1'(t)$ and $d\_{11}'$, which latter is a rational function, which is easily seen to be $<0$. So, $d\_{11}$ decreases. Also, $d\_{11}(0+)=0$. So, $d\_{11}<0$ (on $(0,1]$) and hence $d\_1$ decreases. Also, $d\_1(0+)=0$. So, $d\_1<0$, which yields (2).


To prove (3), rewrite it as  

\begin{equation\*}
 d\_2(t):=\ln \left(\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}
 \left(\frac{1}{t}\right)^{\frac{1}{t}}\right)-\ln \left(\frac{e (2-t)}{2 t}\right)\le0 
\end{equation\*}
(for $t\in(0,1]$).
Consider then $d\_{21}(t):=t^2d\_2'(t)$ and $d\_{21}'$, which latter is a rational function, which is easily seen to be $<0$ on $(0,1)$. So, $d\_{21}$ decreases. Also, $d\_{21}(0+)=0$. So, $d\_{21}<0$ (on $(0,1]$) and hence $d\_2$ decreases. Also, $d\_2(0+)=0$. So, $d\_2<0$, which yields (3).


Finally, (4) is elementary, since $g\_1g\_2$ is a polynomial (of degree $3$). $\Box$


