Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$

Question

Define $a_{n}=1!\left( 1^{1}+1+1\right) +2!\left( 2^{2}+2+1\right) +3!\left( 3^{2}+3+1\right) +\ldots +n!\left( n^{2}+n+1\right) $ , Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$

Answer 1

The trick is to find a useful simplification of $a_n$. This takes a certain amount of inspiration. It’s not obvious from the first few values; you might spot it from the first few values of $1+a_n$, but then again you might not. It’s tempting to rewrite $k^2+k+1$ as $\frac{k^3-1}{k-1}$, but that doesn’t immediately suggest any further simplification, so you might try to do something directly with $k!(k^2+k+1)$. Since $k^2+k=k(k+1)$, and the factor of $k+1$ combines nicely with $k!$, you might try that:

$$k!(k^2+k+1)=k(k+1)k!+k!=k(k+1)!+k!=k\Big((k+1)!+(k-1)!\Big)\;.$$

Then

$$\begin{align*} a_n&=\sum_{k=1}^nk\Big((k+1)!+(k-1)!\Big)\\\\ &=\sum_{k=1}^nk(k-1)!+\sum_{k=1}^nk(k+1)!\\\\ &=\sum_{k=0}^{n-1}(k+1)k!+\sum_{k=2}^{n+1}(k-1)k!\\\\ &=1+2+\sum_{k=2}^{n-1}(k+1)k!+\sum_{k=2}^{n-1}(k-1)k!+(n-1)n!+n(n+1)!\\\\ &=3+\sum_{k=2}^{n-1}\Big((k+1)+(k-1)\Big)k!+(n-1)n!+n(n+1)!\\\\ &=3+2\sum_{k=2}^{n-1}kk!+(n-1)n!+n(n+1)!\\\\ &=3+2\sum_{k=2}^{n-1}\Big((k+1)k!-k!\Big)+(n-1)n!+n(n+1)!\\\\ &=3+2\sum_{k=2}^{n-1}\Big((k+1)!-k!\Big)+(n-1)n!+n(n+1)!\\\\ &=3+2(n!-2!)+(n-1)n!+n(n+1)!\\\\ &=(n+1)n!+n(n+1)!-1\\\\ &=(n+1)(n+1)!-1\;, \end{align*}$$

so $1+a_n=(n+1)(n+1)!$. Now get a closed form for $\sum_{k=1}^n2k$, simplify, and write the series as a telescoping series.

Answer 2

Observe that the $r$th term $(1\le r\le n)$ of $a_n$ is $\displaystyle(r^2+r+1)\cdot r!$

Now as $r^2+r+1$ is Quadratic, we can write $\displaystyle r^2+r+1=(r+2)(r+1)+A(r+1)+B$ where $A,B$ are arbitrary constants, so that

$\displaystyle (r^2+r+1)\cdot r!=\{(r+2)(r+1)+A(r+1)+B\}\cdot r!$ $\displaystyle=(r+2)!+A\cdot (r+1)! +B\cdot r!$

If it were a cubic polynomial, we had to take $\displaystyle(r+3)(r+2)(r+1)+A(r+2)(r+1)+B(r+1)+C$ and so on.

Anyway, let us determine $A,B:$

we have $\displaystyle r^2+r+1=(r+2)(r+1)+A(r+1)+B$ $\displaystyle\implies r^2+r+1=r^2+r(3+A)+2+A+B$

Comparing the coefficients of $\displaystyle r,3+A=1\implies A=-2$

Comparing the constants $\displaystyle 2+A+B=1\implies B=1$

$\displaystyle\implies r!(r^2+r+1)=(r+2)!- 2(r+1)!+r!$ $\displaystyle=\underbrace{(r+2)!- (r+1)!}-\underbrace{\{(r+1)!-r!\}}$

Observe the Telescoping Series

Set $\displaystyle r=1,2,\cdots,n$ to find

$\displaystyle a_n=(n+2)!-2!-\{(n+1)!-1\}=(n+1)!(n+2-1)-1$ $\displaystyle=(n+1)\cdot(n+1)!-1$

As Cocopuffs has derived each summand to be $\displaystyle=\frac{n(n+1)}{(n+1)\cdot(n+1)!}=\frac n{(n+1)!}$

$$\text{But, }\frac n{(n+1)!}=\frac {n+1-1}{(n+1)!}=\frac1{n!}-\frac1{(n+1)!}$$

Again, another Telescoping Series

Answer 3

You can show that $1+a_n = (n+1) \cdot (n+1)!$ by induction on $n$.

Since $2+4+6+ ... + 2n = n(n+1)$, your sum becomes $$\sum_{n=2}^{\infty} \frac{n}{(n+1)!} = \sum_{n=2}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{(n+1)!} = \frac{1}{2}.$$