Vtyjkyuk

Consider the following series,

$$1+frac1+22!+frac1+2+33!+cdots$$

My Efforts

The series can be written in compact for as,

$$sum_n=1^inftyn(n+1)over 2n!$$

$$=1over 2(sum_n=1^inftyn^2over n!+sum_n=1^inftyn over n!)$$
$$=1over 2(sum_n=1^inftyn^2over n!+sum_n=1^infty1 over (n-1)!)$$

Now
$$sum_n=1^infty1 over (n-1)!=1+1+1over 2!+1over 3!+cdots=e$$
First summand can be simplified as,
$$sum_n=1^inftyn over (n-1)!$$

But where does it converge? Any hints?

Edit: Thank you for all the quick replies. I now realize it was so easy. I just had to use $n=n-1+1$.

Hint:

$$
sum_n=1^inftyn over (n-1)! = sum_n=1^inftyn - 1 + 1 over (n-1)!
$$

$$
= sum_n=2^infty1 over (n-2)! + sum_n=1^infty 1 over (n-1)!
$$

beginalign
sum_n=1^inftyn(n+1)over 2n!
&= sum_n=1^inftyn-1+2over 2(n-1)! \
&= sum_n=2^inftydfrac12(n-2)!+sum_n=1^inftydfrac1(n-1)! \
&= dfrac12e+e \
&= colorbluedfrac32e
endalign

Hint: $sum_k=1 fracn^2n! x^n =e^x (x^2 + x)$.

This can easily be proven by considering the Taylor expansion of $e^x$, multiplying, and making the change of variable.

To me, the series is summarized as $$sum_1^inftyfracn+12(n-1)!$$ instead! Now, use the ratio test.

HInt: when $ngeq 2$, $$fracn (n-1)!=fracn-1+1(n-1)!=frac1(n-2)!+frac1(n-1)!$$

$sum_n=1^inftyfracnleft(n-1right)!=1+sum_n=2^inftyfracleft(n-1right)+1left(n-1right)!=1+sum_n=0^inftyfrac1n!+sum_n=1^inftyfrac1n!=1+e+(e-1)=2e$