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Hi I'm self studying calculus, I tried wolframalpha and it says it converges but the final form looks odd, I'm hoping that it converges to a simpler form like what $$sum_{n=0}^{infty}frac{na^n}{n!} = ae^a$$ converges to according to wikipedia. What does $sum_{n=M}^{infty}frac{na^n}{n!}$ converges to if it's offset?

It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.

In fact your sum is

$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$

where $x$ is your $M$, but can be also a real or complex number.

-- Addendum --

Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.

The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)

The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$

Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.

Also refer to this Introduction given at Wolfram Functions Site.

So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)

For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.

It converges to

$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$

for example

$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$

$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$

I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$

I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).