series in sigma notation, find the sum

**kcsteven** · February 4th 2007, 05:34 AM

I have been sitting and reading two different text books and following examples but I still can not figure out what I am doing. I am trying to write each series in sigma notation and find its sum. Here are a few examples I am working on. 1+1/2+1/2^2+...+1/2^60, 1+1/2+1/3+...+1/100, 1+3+5+...+201, 2-2^3+2^5-2^7+...-2^15
The text explains sigma notation and I understand the lower and upper limit but I am having a hard time taking the information in the book, putting it into my brain and having something logical comming out that gives me the answer. I worked out the examples for the problems with the sum of a series but there were not any examples that looked like these, so I could not look at how they did one of these. It always helps when there is an example and it is broken down step by step so one can see how they arrived at the answer. I would appreciate any and all examples so I can grasp this concpet and finish the rest of my homework.
Thank you!!!
Keith Stevens

**ThePerfectHacker** · February 4th 2007, 05:44 AM

Originally Posted by kcsteven

1+1/2+1/2^2+...+1/2^60

We can write,
$\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\fra c{1}{2^{60}}$
Thus, the exponent is chaning from $0$ to $60$ .
In sigma notation,
$\sum_{k=0}^{60} \frac{1}{2^k}$
Because everything else remains fixed.
The sum of this series is geometric.
Use the geometric series formula with constant ratio $1/2$ .
$\frac{1-(1/2)^{60+1}}{1-1/2}$

1+1/2+1/3+...+1/100

You can write,
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1 00}$
Only the denominator is changing from $1$ to $100$ .
In sigma notation,
$\sum_{k=1}^{100} \frac{1}{k}$
There is no nice formula for this series, but we can approximate it with the natural logarithm.
Approximately the value is $\ln 100\approx 4.6$ .

**kcsteven** · February 4th 2007, 05:52 AM

Thanx alot!!! I got it!!!!!!!!!!!!!!!!!!!
I don't understand why sometimes I can sit and try the problems over and over again and not get the answer but when shown one more example It becomes clear, go figure. Anyway I really appreciate the example It did the trick!!!!
Thank you again,
Keith Stevens

**ThePerfectHacker** · February 4th 2007, 05:57 AM

Originally Posted by kcsteven

1+3+5+...+201

This is an arithmetic series, each term increases by 2.
Thus, in general the formula for each term is,
$an+b$ where $n$ is the n-th position.
And where $a,b$ are certain numbers which we need to determine.
Note, that when $n=1$ we have the first term is 1 also thus,
$a(1)+b=1$
And when $n=2$ we have 3.
Thus,
$a(2)+b=3$
Solve the system of linear equation,
$a=2$ and $b=-1$ .
Thus, the formula for the n-th term is,
$2n-1$ .

Now what term is 201?
We need to solve,
$2n-1=201$
$n=101$ .
Thus, the pattern is this,
$2(1)-1,2(2)-1,2(3)-1,....,2(101)-1$ .
In sigma notation,
$\sum_{k=1}^{101}(2k-1)=\sum_{k=1}^{101}2k-\sum_{k=1}^{101}1=2\sum_{k=1}^{101}k-\sum_{k=1}^{101}1$
Using the formula,
$1+2+...+n=\sum_{k=1}^n k =\frac{n(n+1)}{2}$
We have,
$2\left( \frac{101(102)}{2}\right) - 101$
Becuase,
$\sum_{k=1}^{101}1=1+1+...+1=101$
Thus,
$101(102)-101=101(102-1)=101(101)=101^2=10201$

2-2^3+2^5-2^7+...-2^15

Whenever you have an alternating sequence always consider $(-1)^k$ because this alternates.
Thus,
$(-1)^02^{2\cdot1-1},(-1)^12^{2\cdot2-1},(-1)^22^{2\cdot 3-1},...,(-1)^72^{2\cdot 8-1}$
Thus,
$\sum_{k=1}^8 (-1)^{k-1}2^{2k-1}$

**kcsteven** · February 4th 2007, 06:57 AM

Thankx, for the examples, I am crusing thru these problems now. I appreciate your help because I would have had to wait until Monday to ask my questions, now I will not feel like I am falling behind.

Thank You!!!!!!!
Keith Stevens

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