Simplifcation of a sum I’m unsure if I’m allowed to simplify like this:
$$\begin{align*}
\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} &=  \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7^{n} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7}{10}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{10}
\end{align*}$$
Please tell me whether this is right or wrong and if possible how else the manipulated part could be simplified. Thanks in advance!

Your first two steps are correct, but then you went astray.
$$\sum_{n=2}^\infty\frac{7^n(n-1)!}{10^n(n-1)!}=\sum_{n=2}^\infty\frac{7^n}{10^n}=\sum_{n=2}^\infty\left(\frac7{10}\right)^n\;,$$
not $\sum_{n=2}^\infty\frac7{10}$. Now
$$\frac17\sum_{n=2}^\infty\left(\frac7{10}\right)^n=\frac17\cdot\frac{\left(\frac7{10}\right)^2}{1-\frac7{10}}=\frac17\cdot\frac{10}3\cdot\frac7{10}\cdot\frac7{10}=\frac7{30}\;,$$
since the sum is just a geometric series. And you should note that $\sum_{n=2}^\infty\frac7{10}$ diverges, so $\frac17\sum_{n=2}^\infty\frac7{10}$ definitely isn’t $\frac1{10}$ even if this were the right expression to be evaluating.
The first summation can also be simplified. For starters,
$$\sum_{n=2}^\infty\frac{2^{n+3}}{10^n(n-1)!}=8\sum_{n=2}^\infty\frac{2^n}{10^n(n-1)!}=8\sum_{n=2}^\infty\left(\frac15\right)^n\frac1{(n-1)!}\;;$$
this at least gives the two exponentials the same exponent, so that they can be combined into a single exponential. But we can do better if we remember that
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\;.\tag{1}$$
We first pull out a factor of $\frac15$ to make the exponent on $\frac15$ match the factorial:
$$8\sum_{n=2}^\infty\frac{2^n}{10^n(n-1)!}=8\sum_{n=2}^\infty\left(\frac15\right)^n\frac1{(n-1)!}=\frac85\sum_{n=2}^\infty\left(\frac15\right)^{n-1}\frac1{(n-1)!}\;.$$
Now shift the index of summation one place: as $n$ runs up from $2$, $n-1$ runs up from $1$, and
$$\frac85\sum_{n=2}^\infty\left(\frac15\right)^{n-1}\frac1{(n-1)!}=\frac85\sum_{n=1}^\infty\frac{(1/5)^n}{n!}\;.$$
Now use $(1)$ to express this in closed form, i.e., without a summation.
