Chapter 12- Section 4(Advanced Mathematical concepts)
Convergent and Divergent Series
ICM
By Arial H
Using the ratio test to determine whether the following series are convergent or divergent:
| 2/7+4/8+6/9+8/10+... or could be seen like this: 2/7+1/2+2/3+4/5+... | First
find the nth term |
| The nth term is what n has to equal to become the next term in the series. | The nth term = 2n/n+5 |
Now in the nth
term wherever there is an n you have to substitute n-1 (n=n-1)
because r=lim/n->oo=an-1/an
|
r=lim/n->2n-1/(n-1)+5 or 2n-1/ n+4 |
| The next step is to put the new equation over the original equation and began to find the limit. (n->oo) | r=lim/n->oo(2n-1/n+4)/(2n/n+4) |
| to solve multiply by the reciprocal of the denominator. | r=lim/n->oo (2n-1/n+4 )*( n+4/2n) |
| r=lim/n->oo (4n2-2n)/(n2+8n+16) | This is the answer after multiplying the reciprocal |
| Now you want to divide by the highest power in the denominator, which is n2 | r=lim/n->oo(4n2-2n/n2)/(n2+8n+16/n2)= r=lim/n->oo(4-2/n)/(1+8/n+16/n2) |
| It is known that when ever a number is divided by n the value is constantly getting smaller and closer to 0, so the result of that in this problem is: | r=lim/n->oo(4-0)/(1+0+0)= (4)/(1)=(4) |
r=
4, this you now know. The next thing you need to know is how to determine
whether the series is divergent or convergent using the r.
|
if r<1 the series is convergent, if r is >1 the series is divergent and if r=1 the test provides no information. |
| so according to the ratio test the series is DIVERGENT!!!! |
*pictures were received from: www.bellsnwhistles.com
Quote: "Learn from yesterday, live for today, hope for tomorrow."
-Anonymous
This quotation means that take everyday day by day and take something out of that day to use for the next day and those days to follow. Persevere to make everyday worth while.
For more information on this topic refer to:
http://math.furman.edu/~dcs/book/c5pdfsec54.pdf