How to prove $\sum _ { n = 1 } ^ { \infty } \frac { n a ^ { n - 1 } } { ( n - 1 ) ! } = ( a + 1 ) e ^ { a }$ and $\sum _ { n = 1 } ^ { \infty } \frac { n ^ { 2 } a ^ { n - 1 } } { ( n - 1 ) ! } = ( a ^ { 2 } + 3 a + 1 ) e ^ { a }$?
I know $\sum _ { n = 1 } ^ { \infty } \frac { a ^ { n - 1 } } { ( n - 1 ) ! } = e ^ { a }$ using Taylor's Formula. How can I apply it to the above two?
$$ae^a=\sum_{n=1}^{\infty}\frac{a^n}{(n-1)!}$$ differentiate $$ae^a+e^a=\sum_{n=1}^{\infty}\frac{na^{n-1}}{(n-1)!}$$ multiply $a$ on bs $$a^2e^a+ae^a=\sum_{n=1}^{\infty}\frac{na^n}{(n-1)!}$$ Now diffrentiate again what do you get??
Take the derivaty:
$\sum _ { n = 1 } ^ { \infty } \frac { (n-1)a ^ { n -2} } { ( n - 1 ) ! } = e ^ { a }$
multiply it by $a$:
$\sum _ { n = 1 } ^ { \infty } \frac { (n-1)a ^ { n -1} } { ( n - 1 ) ! } = a \cdot e ^ { a }$
distribute in $(n-1)$ and pass the term multiplied in the $1$ to the other side and you will get your answer.
Same principle works for second equality, take the derivative and isolate the $n^2$
Like Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$
for $n\ge2,$
$$\dfrac n{(n-1)!}=\dfrac{1+n-1}{(n-1)!}=\dfrac1{(n-1)!}+\dfrac1{(n-2)!}$$
$$\implies\sum_{n=1}^\infty\dfrac{na^{n-1}}{(n-1)!}=\sum_{n=1}^\infty\dfrac{a^{n-1}}{(n-1)!}+\sum_{n=2}^\infty\dfrac{a^{n-1}}{(n-2)!}$$
Set $n-1=m$ to find the first infinite sum $ =e^a$
The second sum $$=a\sum_{n=2}^\infty\dfrac{a^{n-2}}{(n-2)!}=ae^a$$
Similarly for $n^2,$ write $$n^2=(n-1)(n-2)+a(n-1)+b$$ so that for $n-3\ge0,$
$$\dfrac{n^2}{(n-1)!}=\dfrac1{(n-3)!}+\dfrac a{(n-2)!}+\dfrac b{(n-1)!}$$
Now comparing the coefficients of $n,$ $$0=-a+3\iff a=3$$
Compare the constants, $0=b-a+2\iff b=a-2=?$
Similarly for $n^3,$ write $$n^3=(n-1)(n-2)(n-3)+c(n-1)(n-2)+d(n-1)+e$$
Generally to do these kind of problems see how to map from the first summation to second,
$$ \sum_{n=1}^{\infty} \frac{na^{n-1} }{(n-1)!} \xrightarrow[]{?} \sum_{n=1}^{\infty} \frac{n^2 a^{n-1}}{(n-1)!}$$
We can find by inspection that the operation we have to do is:
$$ Q() = \frac{d}{da} \left[a \left( \right)\right]$$
i.e: multiply by 'a' then differentiate. Consider our original series,
$$ \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!}= (a+1)e^a$$
Apply $Q$ on both sides:
$$ Q \left( \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!} \right)= \frac{d}{da} \left[ a( a+1) e^a \right]$$
By the premise, we can directly right LHS and simplifying the RHS by triple product rule:
$$ \sum_{n=1}^{\infty} \frac{n^2 a^{n-1}}{(n-1)!} = (a+1) e^a + a e^a + a(a+1) e^a$$
You'll reach the answer after simplifying RHS using algebra :)
Oh, and for deriving the first summation, consider LHS: $$ \sum_{n=1}^{\infty} \frac{na^{n-1}}{(n-1)!}$$
And do a change of variables $n-1 = i$,
$$ \sum_{i=0}^{\infty} \frac{ (i+1)a^i }{(i)!} =\sum_{i=1}^{\infty} \frac{a^i}{(i-1)!} + \sum_{i=0}^{\infty} \frac{a^i}{i!} = a \sum_{i=1}^{\infty} \frac{a^{i-1} }{(i-1)!} + e^a = a e^a + e^a$$
Note:
$$? = \sum_{i=1}^{\infty} \frac{a^{i-1}}{(i-1)!} = \sum_{j=0}^{\infty} \frac{ a^j}{j!}$$
By reindexing
