6.4: Sum of a Series (2024)

To find approximate solutions to problems in the sciences, it is often necessary to calculate the sum of a finite or infinite series. There are a variety of formulas that are used to accomplish this. Some of these formulas will be presented with proofs, but others will not. If you are interested in the proofs that are not included, please let me know.
General Formulas

Constant Series - notice that there is no \(k\) in the summation, the \(c\) is a constant that does not depend on the value of \(k\)
\(\sum_{k=1}^{n} c=c+c+c+\cdots+c=n * c\)

Sum of the first \(n\) integers:
\(\sum_{k=1}^{n} k=1+2+3+\dots+n=\frac{n(n+1)}{2}\)

Sum of the first \(n\) perfect squares:
\(\sum_{k=1}^{n} k^{2}=1+4+9+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\)

Sum of the first \(n\) perfect cubes:
\(\sum_{k=1}^{n} k^{3}=1+8+27+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}\)

The first formula should be obvious. The other three formulas are usually proved using mathematical induction, which we won't cover in this course. If you're interested in these proofs and how mathematical induction works, please let me know.

Formulas for the sum of arithmetic and geometric series:

Arithmetic Series: like an arithmetic sequence, an arithmetic series has a constant difference \(d .\) If we write out the terms of the series:
\(\sum_{k=1}^{n} a_{k}=a_{1}+a_{2}+a_{3}+\cdots+a_{n}\)

we can rewrite this in terms of the first term \(\left(a_{1}\right)\) and the constant difference \(d\)
\(\sum_{k=1}^{n} a_{k}=a_{1}+\left(a_{1}+d\right)+\left(a_{1}+2 d\right)+\cdots+\left(a_{1}+(n-1) d\right)\)

This expression is equivalent to:
\(\sum_{k=1}^{n} a_{k}=\left(a_{1}+a_{1}+a_{1}+\cdots+a_{1}\right)+(d+2 d+3 d+\cdots(n-1) d)\)
\(\sum_{k=1}^{n} a_{k}=n a_{1}+d(1+2+3+\cdots(n-1))\)

Using the previous formula for the sum \(1+2+3+\dots+(n-1)\) gives us:
\(\sum_{k=1}^{n} a_{k}=n a_{1}+d\left(\frac{(n-1) n}{2}\right)\)

This formula is often stated in various forms:
\(\sum_{k=1}^{n} a_{k}=\frac{n}{2}\left(2 a_{1}+(n-1) d\right)\)

Given a geometric series, whose first term is \(a\) and with a constant ratio of \(r\) \(\sum_{k=1}^{n} a * r^{k-1},\) we can write out the terms of the series in a similar way that we did for the arithmetic series.
\(\sum_{k=1}^{n} a * r^{k-1}=a+a r+a r^{2}+a r^{3}+\cdots+a r^{n-1}\)

The trick to finding a formula for the sum of this type of series is to multiply both sides of the previous equation by \(r\)
For simplicity's sake let's rename the sum of the series \(\sum_{k=1}^{n} a * r^{k-1}\) as \(S_{n}\)

So,
\(S_{n}=a+a r+a r^{2}+a r^{3}+\dots+a r^{n-1}\)

and
\(r * S_{n}=r\left(a+a r+a r^{2}+a r^{3}+\dots+a r^{n-1}\right)=a r+a r^{2}+a r^{3}+\dots+a r^{n}\)

If we subtract these two equations, we will have:
\(S_{n}=a+a r+a r^{2}+a r^{3}+\dots+a r^{n-1}\)
\(-\left(r * S_{n}=a r+a r^{2}+a r^{3}+\cdots+a r^{n}\right)\)

Then we'll have:
\(S_{n}-r S_{n}=a-a r^{n}\)

Factor out \(S_{n}\) on the left-hand side:
\(S_{n}(1-r)=a-a r^{n}\)

and divide on both sides to isolate \(S_{n}:\)
\(\frac{S_{n}(1-r)}{(1-r)}=\frac{a-a r^{n}}{(1-r)}\)
\(S_{n}=\frac{a-a r^{n}}{1-r}\)

So for a finite geometric series, we can use this formula to find the sum. This formula can also be used to help find the sum of an infinite geometric series, if the series converges. Typically this will be when the value of \(r\) is between -1 and 1. In other words, \(|r|<1\) or \(-1<r<1 .\) This is important because it causes the \(a r^{n}\) term in the above formula to approach 0 as \(n\) becomes infinite. So, if \(-1<r<1,\) then the sum of an infinite geometric series wil be:
\(S_{n}=\frac{a}{1-r}\)

Exercises 6.4
Find the sum for each of the following finite geometric series.
1) \(\sum_{k=1}^{7} 3\left(\frac{1}{4}\right)^{k-1}\)

2) \(\sum_{k=1}^{7} 16\left(\frac{1}{3}\right)^{k-1}\)

3) \(\sum_{k=1}^{7} 3^{k}\)

4) \(\sum_{k=1}^{10} 2^{k-1}\)

5) \(\sum_{k=1}^{5} 4^{k-1}\)

6) \(\sum_{k=1}^{4} 6^{k-1}\)

7) \(\sum_{k=1}^{7} 2^{k}\)

8) \(\sum_{k=1}^{8} 3^{k}\)

9) \(\sum_{k=1}^{5} 2^{k+2}\)

10) \(\sum_{1}^{6} 3^{k-4}\)

Determine whether each of the following geometric series has a sum. If it does, use the formula \(S_{n}=\frac{a}{1-r}\) to find the sum.
11) \(\sum_{k=1}^{\infty} 5 *\left(\frac{2}{3}\right)^{k-1}\)

12) \(\sum_{k=1}^{\infty} 12 *\left(\frac{1}{2}\right)^{k-1}\)

13) \(\sum_{k=1}^{\infty}\left(\frac{3}{4}\right)^{k-1}\)

14) \(\sum_{k=1}^{\infty}\left(\frac{3}{5}\right)^{k-1}\)

15) \(\sum_{k=1}^{\infty}\left(\frac{1}{5}\right)^{k+1}\)

16) \(\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k+1}\)

17) \(\sum_{k=1}^{\infty}\left(\frac{4}{3}\right)^{k-1}\)

18) \(\sum_{k=1}^{\infty}\left(\frac{3}{2}\right)^{k-1}\)

19) \(\sum_{k=1}^{\infty}\left(-\frac{1}{3}\right)^{k+2}\)

20) \(\sum_{k=1}^{\infty}\left(-\frac{1}{2}\right)^{k+4}\)