# Evaluating Series

We already learned about arithmetic, geometric as well as telescoping series. Today let’s learn how to evaluate series. If you’re thinking to create your own formula, this is the perfect topic for you. Before I teach you to create your own formula and you’re so lucky if this is your first time. We need to know first some basic information. One of them is to know what summation symbol is all about.

Short Hand                                       Expanded

$\displaystyle\sum_{k=1}^{5} k$                                          $1+2+\cdots+5$

$\displaystyle\sum_{k=1}^{n} k^2$                                      $1^2+2^2+\cdots+n^2$

$\displaystyle\sum_{k=2}^{n} k^3$                                      $2^3+3^3+\cdots+n^3$

Foundation Formula:

$\displaystyle\sum_{k=1}^n k=\displaystyle\frac{n(n+1)}{2}$     (1)

$\displaystyle\sum_{k=1}^n k^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$      (2)

$\displaystyle\sum_{k=1}^n k^3=[\displaystyle\frac{n(n+1)}{2}]^2$       (3)

For all positive integer a and $a>1$

$\displaystyle\sum_{k=1}^\infty \displaystyle\frac{1}{a^k}=\displaystyle\frac{1}{a-1}$        (4)

$\displaystyle\sum_{k=1}^\infty \displaystyle\frac{k}{a^k}=\displaystyle\frac{a}{(a-1)^2}$     (5)

$\displaystyle\sum_{k=1}^\infty \displaystyle\frac{k^2}{a^k}=\displaystyle\frac{a(a+1)}{(a-1)^3}$  (6)

Some Rules to follow:

$\displaystyle\sum cu=c\sum u$  where $c$ is constant and $u$ is a function.       (i)

$\displaystyle\sum (u\pm v)=\sum u\pm \sum v$     (ii)

Sample Problem 1:

Solve the formula for the sum of  $1\cdot 2+2\cdot 3+\ldots+n(n+1)$

Step 1: Write the expression to its shorthand notation using the summation symbol

$\sum_{n=1}^n n(n+1)$

Step 2: Evaluate the series following the rules and use the basic foundation of series

$\displaystyle\sum_{n=1}^n n(n+1)\leftrightarrow \sum_{n=1}^n (n^2+n)\leftrightarrow \displaystyle\sum_{n=1}^n n^2 +\displaystyle\sum_{n=1}^n n$

But $\displaystyle\sum_{k=1}^n k=\displaystyle\frac{n(n+1)}{2}$  and  $\displaystyle\sum_{k=1}^n k^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

$\displaystyle\sum_{n=1}^n n^2 +\displaystyle\sum_{n=1}^n n=\displaystyle\frac{n(n+1)(2n+1)}{6}+\displaystyle\frac{n(n+1)}{2}$

$=\displaystyle\frac{n(n+1)}{2}[\displaystyle\frac{2n+1}{2}+1]$

$=\displaystyle\frac{n(n+1)}{2}[\displaystyle\frac{2n+4}{3}]$

$=\displaystyle\frac{n(n+1)(n+2)}{3}$

Thus,

$\displaystyle\sum_{n=1}^n n(n+1)= \boxed{\displaystyle\frac{n(n+1)(n+2)}{3}}$

Sample Problem 2:

Find the sum of series

$1\cdot 2^2+2\cdot 3^2+\ldots+99\cdot 100^2$

Solution:

We need to find the first the formula for the nth term of the series. It is obvious that the nth term can be express as $n(n+1)^2$.

In shorthand notation we can express the sum as

$\displaystyle\sum_{n=1}^n n(n+1)^2$

Simplifying the notation we have

$=\displaystyle\sum_{n=1}^n (n^3+2n^2+n)$

Using rule (i) & (ii)

$\displaystyle\sum_{n=1}^n (n^3+2n^2+n)=\displaystyle\sum_{n=1}^n n^3 +\displaystyle\sum_{n=1}^n 2n^2+\sum_{n=1}^n n$

$\displaystyle\sum_{n=1}^n (n^3+2n^2+n)=\displaystyle\sum_{n=1}^n n^3+ 2\displaystyle\sum_{n=1}^n n^2+\displaystyle\sum_{n=1}^n n$

Using formula (3),(2), and (1) we have

$\displaystyle\sum_{n=1}^n n^3 +2\displaystyle\sum_{n=1}^n n^2+\displaystyle\sum_{n=1}^n n$

$(\displaystyle\frac{n(n+1)}{2})^2+2\displaystyle\frac{n(n+1)(2n+1)}{6}+\displaystyle\frac{n(n+1)}{2}$

But   $\displaystyle\frac{n(n+1)}{2}$   is a common factor

$\displaystyle\frac{n(n+1)}{2}[\displaystyle\frac{n(n+1)}{2}+\displaystyle\frac{2(2n+1)}{3}+1]$

$\displaystyle\frac{n(n+1)}{2}[\displaystyle\frac{3(n)(n+1)+4(2n+1)+6}{6}]$

$\displaystyle\frac{n(n+1)}{2}[\displaystyle\frac{3n^2+11n+10}{6}]$

By factoring,  $3n^2+11n+10=(3n+5)(n+2)$

$\displaystyle\frac{n(n+1)(n+2)(3n+5)}{12}$

Hence,

$\displaystyle\sum_{n=1}^n n(n+1)^2=\boxed{\displaystyle\frac{n(n+1)(n+2)(3n+5)}{12}}$

Going back to the problem we have $n=99$

$\displaystyle\sum_{n=1}^{99} n(n+1)^2=\displaystyle\frac{99(99+1)(99+2)(3(99)+5)}{12}$

$\displaystyle\sum_{n=1}^{99} n(n+1)^2=\boxed{25,164,150}$

Sample Problem 3:

Find the sum of the infinite series

$\displaystyle\frac{1}{6}+\displaystyle\frac{2}{6^2}+\displaystyle\frac{3}{6^3}+\ldots$

Expressing the sum in shorthand notation we have

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{6^n}$

This notation is similar to formula (5) with $a=6$

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{a^n}=\displaystyle\frac{a}{(a-1)^2}$

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{6^n}=\displaystyle\frac{6}{(6-1)^2}$

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{6^n}=\boxed{\displaystyle\frac{6}{25}}$

Sample Problem 4:

Find the sum of infinite series

$\displaystyle\frac{1}{5}+\displaystyle\frac{4}{25}+\displaystyle\frac{9}{125}+\ldots$

Solution:  We can express the series in shorthand notation as follows.

$\displaystyle\sum_{n=1}^n \displaystyle\frac{n^2}{a^n}$

The notation above is similar to formula (6) where $a=5$

$\displaystyle\sum_{n=1}^n \displaystyle\frac{n^2}{a^n}=\displaystyle\frac{a(a+1)}{(a-1)^3}$

$\displaystyle\sum_{n=1}^n \displaystyle\frac{n^2}{5^n}=\displaystyle\frac{5(5+1)}{(5-1)^3}$

$\displaystyle\sum_{n=1}^n \displaystyle\frac{n^2}{5^n}=\boxed{\displaystyle\frac{15}{32}}$

Sample Problem 5:

Find the sum of infinite series

$\displaystyle\frac{1\cdot 3}{3}+\displaystyle\frac{2\cdot 4}{3^2}+\displaystyle\frac{3\cdot 5}{3^3}+\ldots$

Solution: We can express the sum in shorthand notation as follows

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n(n+2)}{3^n}$

Using the rule (ii)

$\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n(n+2)}{3^n}=\sum_{n=1}^\infty \displaystyle\frac{n^2+2n}{3^n}$

$=\displaystyle\sum_{n=1}^\infty (\displaystyle\frac{n^2}{3^n}+\displaystyle\frac{2n}{3^n})$

$=\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n^2}{3^n}+2\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{3^n}$

The two sums are similar to (6) and (5) respectively where $a=3$

$=\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n^2}{3^n}+2\displaystyle\sum_{n=1}^\infty \displaystyle\frac{n}{3^n}$

$=\displaystyle\frac{a(a+1)}{(a-1)^3}+2\displaystyle\frac{a}{(a-1)^2}$

$=\displaystyle\frac{3(3+1)}{(3-1)^3}+2\displaystyle\frac{3}{(3-1)^2}$

$=\boxed{3}$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.