You could use perturbation method to calculate this sum. Let's start from:
\begin{gathered}S_n=\sum\limits_{k=0}^nk!(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}\end{gathered}
On the other hand, we have:
\begin{gathered}S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!==1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}\end{gathered}
S
n+1
=
k=0
β
n+1
k!=0!+
k=1
β
n+1
k!=1+
k=1
β
n+1
k!=1+
k=0
β
n
(k+1)!=
=1+
k=0
β
n
k!(k+1)=1+
k=0
β
n
(kβ
k!+k!)=1+
k=0
β
n
kβ
k!+
k=0
β
n
k!
(2)
S
n+1
=1+
k=0
β
n
kβ
k!+S
n
So from (1) and (2) we have:
\begin{gathered}\begin{cases}S_{n+1}=S_n+(n+1)!S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases} S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n (\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}\end{gathered}
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S
n+1
=S
n
+(n+1)!
S
n+1
=1+
k=0
β
n
kβ
k!+S
n
S
n
+(n+1)!=1+
k=0
β
n
kβ
k!+S
n
(β)
k=0
β
n
kβ
k!=(n+1)!β1
Now, let's try to calculate sum \sum\limits_{k=0}^{n}k\cdot k!
k=0
β
n
kβ
k! , but this time we use perturbation method.
\begin{gathered}S_n=\sum\limits_{k=0}^nk\cdot k! \boxed{S_{n+1}=S_n+(n+1)(n+1)!} \end{gathered}
S
n
=
k=0
β
n
kβ
k!
S
n+1
=S
n
+(n+1)(n+1)!
but:
\begin{gathered}S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!== \sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!== \sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!==\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n \boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}\end{gathered}
S
n+1
=
k=0
β
n+1
kβ
k!=0β
0!+
k=1
β
n+1
kβ
k!=0+
k=0
β
n
(k+1)(k+1)!=
=
k=0
β
n
(k+1)(k+1)k!=
k=0
β
n
(k
2
+2k+1)k!=
=
k=0
β
n
[(k
2
+1)k!+2kβ
k!]=
k=0
β
n
(k
2
+1)k!+
k=0
β
n
2kβ
k!=
=
k=0
β
n
(k
2
+1)k!+2
k=0
β
n
kβ
k!=
k=0
β
n
(k
2
+1)k!+2S
n
S
n+1
=
k=0
β
n
(k
2
+1)k!+2S
n
When we join both equation there will be:
\begin{gathered}\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases} S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n \sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n== (n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]==(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1== n(n+1)!+1\end{gathered}
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S
n+1
=S
n
+(n+1)(n+1)!
S
n+1
=
k=0
β
n
(k
2
+1)k!+2S
n
S
n
+(n+1)(n+1)!=
k=0
β
n
(k
2
+1)k!+2S
n
k=0
β
n
(k
2
+1)k!=S
n
β2S
n
+(n+1)(n+1)!=(n+1)(n+1)!βS
n
=
=(n+1)(n+1)!β
k=0
β
n
kβ
k!
=
(β)
(n+1)(n+1)!β[(n+1)!β1]=
=(n+1)(n+1)!β(n+1)!+1=(n+1)!β
[n+1β1]+1=
=n(n+1)!+1
So the answer is:
\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}
k=0
β
n
(1+k
2
)k!=n(n+1)!+1
Sorry for my bad english, but i hope it won't be a big problem :)
