Is there a method to sum this infinite series with a lower limit of -1 that decomposes into a geometric and arithmetic series? (partial sum formula) I was presented with the series
$$\sum_{k=-1}^\infty \frac{1}{2^k} + k $$
and asked to write a partial sum formula (in Calculus 3).
I know index shifting this series is easily possible so k = 1, but this was a one-shot stop test problem and so I opted to preserve the original series.
I separated this series into its two components (because I wasn’t able to write a partial sum formula for the original series)
$$\sum_{k=-1}^\infty \frac{1}{2^k} $$ plus $$\sum_{k=-1}^\infty k $$
and I was able to compose a partial sum formula for the first one, although it was definitely challenging. I subtracted a function from the finite limit of the infinite series.
But the other series consisting of just k I am struggling to write a partial sum formula for, but most importantly, am starting to believe that the partial sum formulas of these two series cannot be added back together.
How does one compose a partial sum formula for this?
 A: Rather than shifting the index, it's probably easier to separate the first term, which equals 1. Now you want a partial sum, so we can start with: $$\sum_{k=-1}^n \left(\frac{1}{2^k}+k\right)= 1+\left(\sum_{k=0}^n \frac{1}{2^k}\right)+\left(\sum_{k=0}^n k\right)$$
You said you already know how to do the geometric partial sum, so that leaves the arithmetic one, which expands to: $$\sum_{k=0}^n k = 0+1+2+\cdots +n$$ This is the $n$th triangular number, and its sum is equal to $\frac{n(n+1)}{2}$. You can prove this formula by various means, such as geometrically, or by using induction.
As far as reassembling the pieces into a partial sum for the original series, there's nothing more to do than write down the individual pieces with a plus sign between them.

If you wish to avoid truncating the series, you could write it this way: $$\sum_{k=-1}^n\left(\frac{1}{2^k}+k\right)=\sum_{k=-1}^n\left(2\cdot\frac{1}{2^{k+1}}+(k+1)-1\right)$$ The first term's partial sum is that of a geometric sum with initial term $2$, ratio $\frac12$, and largest power $n+1$; thus: $\frac{2\left(1-\left(\frac12\right)^{n+2}\right)}{1-\frac12}$. The second term's partial sum is the $(n+1)$th triangular number: $\frac{(n+1)(n+2)}{2}$. The last term's partial sum is $-1$ times the number of terms, so: $-(n+2)$.
