# 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

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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 $n+1$.

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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

$$a_n=a_1+(n-1)d$$

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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 $g(m)=f(m-1)$.

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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 $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

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More generally if the numerator is of the form $$a_0+a_1n+a_2n^2+a_3n^3+\cdots,$$ we can express it as $$b_0+b_1n+b_2n(n-1)+b_3n(n-1)(n-2)\cdots$$ –  lab bhattacharjee Mar 29 at 15:12