Series question I was trying this question:
Consider the infinite series
$$\frac{1}{1!} +\frac{4}{2!}+\frac{7}{3!}+\frac{10}{4!}+...$$
If the series continues with the same pattern, find the an expression for the nth term.
I considered the denominator of factorials, which forms an arithmetic series 1,2,3,4.. so with $a=1$ and $d=1$ I established that $u_n = n$ so that the denominator of the nth term would be $n!$, however if I use n = 0, it works fine but then with n = 1 it doesn't work. It only works when n = 1 and so on. 
In sequences and series, is the first term given by n = 0 or n = 1?
Thanks
 A: Observe that the denominator of the $n$th term is $n!$
the numeretor of the $n$th term is the $n$th term of $1,4,7,10,\cdots$ (which is an Arithmetic Series)
i.e., $\displaystyle1+3(n-1)=3n-2$
So, the $n$th term is $\displaystyle\frac{3n-2}{n!}=3\cdot\frac1{(n-1)!}-2\cdot\frac1{n!}$
$$\implies\sum_{n=1}^{\infty}\frac{3n-2}{n!}=3\sum_{n=1}^{\infty}\frac1{(n-1)!}-2\sum_{n=1}^{\infty}\frac1{n!}$$
$$=3\sum_{m=0}^{\infty}\frac1{m!}-2\left(\sum_{n=0}^{\infty}\frac1{n!}-\frac1{0!}\right)$$
Now we know  $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$
A: If the general term of your series is $f(n)$ starting with $n=0$ then you can rewrite it in terms of $m=n+1$ which starts at $m=1$ - the general term $g(m)$ is $g(m)=f(m-1)$.
A: Usually people denote the first term with the index $n=1$, which allows you to say the nth term of the sequence. However, using $n=0$ as the first index is also fairly common, so it is up to you.
You can easily transform a sequence with a starting index of $n=1$ to $n=0$ by replacing every instance of $n$ by $n+1$.
A: Hints:
The denominator advances together with the index: $\;n\to n!\;$ , whether the $\;n$-th numerator is the $\;n$-th element of the arithmetic sequence $\;1,4,7,10,...\;$ .
Further hint: if $\;a_1,a_2,...\;$ is an arithmetic sequence with constant difference $\;d\;$ , then
$$a_n=a_1+(n-1)d$$ 
