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!
 A: 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.
A: $$\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!}+\sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{8}{5^n(n-1)!}+\frac{1}{7}\sum_{n=2}^\infty \left(\frac{7}{10}\right)^n =$$
$$=\frac{8}{5}\sum_{n=2}^\infty \frac{(1/5)^{n-1}}{(n-1)!}+\frac{1}{7}\sum_{n=2}^\infty \left(\frac{7}{10}\right)^n =$$
$$=\frac{8}{5}(e^{1/5}-1)+\frac{1}{7}\left(\frac{1}{1-7/10}-1-\frac{7}{10}\right)$$
