Calculate $\sum_{k=1}^{n} \frac{(-2)^{\underline{k}}}{k}$ I'm having a little trouble checking if this solution to the following problem is correct:
1) Calculate $\sum_{k=1}^{n} \frac{(-2)^{\underline{k}}}{k}$
$$\sum_{k=1}^{n} \frac{(-2)^{\underline{k}}}{k} = \sum_{1}^{n+1} \frac{(-2)^{\underline{k}}}{k}\delta k$$
And on the previous exercise I'm asked to calculate $\Delta (C^{\underline{x}})$ for $C=-2$ and $x=y-2$ Using what was calculated on the even previous exercise: $$\Delta (C^{\underline{x}})=\frac{C^{\underline{x+2}}}{C-x}$$
So, quickly plugged that in:
$$\Delta \left((-2)^{\underline{y-2}}\right)
=
\frac{(-2)^{\underline{y}}}{-2-y+2}=-\frac{(-2)^{\underline{y}}}{y}$$
Therefore $\Delta \left(-(-2)^{\underline{y-2}}\right)
=\frac{(-2)^{\underline{y}}}{y}$ which gives us the solution to the initial sigma $$ \sum_{1}^{n+1} \frac{(-2)^{\underline{k}}}{k}\delta k=-(-2)^{\underline{y-2}}\big |_{1}^{n+1}$$
Is this correct? I'm asking because after trying to get the Delta difference on wolfram, I get something completely different, same by using a sigma calculator I found. Thanks!
 A: Yes, it’s correct. Once you finish the calculation, you find that
$$\sum_{k=1}^n\frac{(-2)^{\underline k}}k=(-2)^{\underline{-1}}-(-2)^{\underline{n-1}}=\frac1{-2+1}+(-1)^n2^{\overline{n-1}}=(-1)^nn!-1\;.\tag{1}$$
If in any doubt, you can now check this simply by proving it by induction on $n$. It’s easy enough to verify that both ends of $(1)$ are $-2$ when $n=1$. Now assume $(1)$. Then
$$\begin{align*}
\sum_{k=1}^{n+1}\frac{(-2)^{\underline k}}k&=(-1)^nn!-1+\frac{(-2)^{\underline{n+1}}}{n+1}\\
&=(-1)^nn!+\frac{(-1)^{n+1}2^{\overline{n+1}}}{n+1}-1\\
&=(-1)^nn!+(-1)^{n+1}\frac{(n+2)!}{n+1}-1\\
&=(-1)^{n+1}\big((n+2)n!-n!\big)-1\\
&=(-1)^{n+1}(n+1)n!-1\\
&=(-1)^{n+1}(n+1)!\;,
\end{align*}$$
as desired. Alternatively, you can check it by evaluating the summation without using finite calculus:
$$\begin{align*}
\sum_{k=1}^n\frac{(-1)^k(k+1)!}k&=\sum_{k=1}^n(-1)^k(k+1)(k-1)!\\
&=\sum_{k=1}^n(-1)^k\big(k!+(k-1)!\big)\\
&=\sum_{k=1}^n(-1)^kk!+\sum_{k=1}^n(-1)^k(k-1)!\\
&=(-1)^nn!+\sum_{k=1}^{n-1}(-1)^kk!-\sum_{k=1}^n(-1)^{k-1}(k-1)!\\
&=(-1)^nn!+\sum_{k=1}^{n-1}(-1)^kk!-\sum_{k=1}^n(-1)^{k-1}(k-1)!\\
&=(-1)^nn!+\sum_{k=1}^{n-1}(-1)^kk!-\sum_{k=0}^{n-1}(-1)^kk!\\
&=(-1)^nn!+\sum_{k=1}^{n-1}(-1)^kk!-\sum_{k=1}^{n-1}(-1)^kk!-1\\
&=(-1)^nn!-1\;.
\end{align*}$$
